What is the quadratic formula?
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
684785
Presuming you are talking about a parabola,\n\nthe vertex is the maximum or minimum points of the parabola, the point at which the parabola curves back towards its original direction. For any parabola, the vertex's x-coordinate is the average of the two solution's coordinates.\n\nthe solution points, know as the root of the parabola, is the point at which the parabola intersects the x axis.\n\nfor the quadratic formula y=ax^2+bx+c,\n\nthe formula b^2-4ac is used to determine the amout of solutions a quadratic equation has. it is also called the discriminant\n\nif the answer is > 0, there are two solutions, or two places where the parabola intersects the x axis, if the answer is < 0, the parabola has NO solutions, it does NOT intersect the x axis, and if the answer is equal to 0, the parabola is tangent, or intersects the x axis at one point, and that point will be the VERTEX.\n\nhoped this helped ^^\n\nand if the
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
85829
I used the quadratic formula x={-b+-sqrootof b^2-4ac}/2a in order to find the factors on both the numerator and denominator\n\nthat leaves {(2x-1)(x-3)}/{4(x-3)(x-3)}\n\nClear (x-3) from both sides, that leaves\n\n(2x-1)/(4(x-3)) your answer
881174
use formulas for cos(2x) and sin(2x) depending on cos(x) and sin (x) and then use the fact that\nsin(2x) = sqrt(1-cos(2x)^2).\nyou will get a quadratic equation that you solve with quadratic formula
530596
Given\nax^2 + bx + c = 0\nDivide all terms by a \nx^2 + (b / a) x + c / a = 0 \nSubtract c / a from both sides \nx^2 + (b / a) x + c / a - c / a = - c / a \nand simplify \nx^2 + (b / a) x = - c / a \nAdd (b / 2a)^2 to both sides \nx^2 + (b / a) x + (b / 2a)^2 = - c / a + (b / 2a)^2 \nto complete the square \n[ x + (b / 2a) ]^2 = - c / a + (b / 2a)^2 \nGroup the two terms on the right side of the equation \n[ x + (b / a) ]^2 = [ b^2 - 4a c ] / ( 4a^2 ) \nSolve by taking the square root \nx + (b / a) = ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nSolve for x to obtain two solutions \nx = - b / a ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nThe term sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } may be written \nsqrt { [ b^2 - 4a c ] / ( 4a^2 ) } = sqrt(b^2 - 4a c) / 2 | a | \nSince 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written \nx = [ -b + sqrt( b^2 - 4a c ) ] / 2 a \nx = [ -b - sqrt( b^2 - 4a c ) ] / 2 a \nThe term b^2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible \ncase 1: If b 2 - 4a c > 0 , the equation has 2 solutions. \ncase 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2. \ncase 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions\nps. I did this as quick as possible, hope everything copied correctly.
225790
Well the answer is 1.5 but you should try working it out for yourself to try and get this answer and you probably need to show your working anyway.\n\nUse this quadratic formula (in the box)\n\nhttp://www.sosmath.com/algebra/factor/fac08/fac08.html
1039684
for x^2+5x-9=0  u can't do this the same way.9 * 1 = 9 but 9 + or - 1 not = 5 and 3 * 3 = 9 but 3 + or - 3 not = 5.\nso there is a formula u got to use here.\nit is  x = {- b + - sqrt(b^2 - 4ac)} / { 2a}\nso from the given quadratic equation,\na = 1,b = 5 and c = 9\nand then u substitute , u will get 2 values of x and that's the answer.remeber we have to use this formula if we are unable to factorise the way u did for the previous one.
883463
Quadratic formula:\nX=(-b+Sqrt(b^2-4ac))/2a\nOR\n(-b-Sqrt(b^2-4ac))/2a\n\nIn case you don't know what a,b,&c stand for, they are the coefficients of the quadratic equation (besides c.)... OK, that was kind of condusing, so I'll just write out the quadratic equation, using a,b,&c where the numbers should be:\nax^2-bx+c\n\nIn your case the equation is:\nx^2 -2x +1=5, which is equal to:\n\n(1)x^6 + (-2)x + (-4)= 0, \n(the correct form for a quadratic equation)\nTherefore, a=1,\n                  b=-2\n                  c=-4\nTherefore,\nX=(-b+Sqrt(b^2-4ac))/2a\nOR\n(-b-Sqrt(b^2-4ac))/2a (The Quadratic Equation)\nbecomes:\n\nX=(2+sqrt(4+16))/2 = (2+sqrt(20))/2 = (2+2sqrt(5))/2 = 1+sqrt(5)\nOR \n1-sqrt(5)
139621
Ran into a little difficulty, I can't find my math editor.\n\nBut here goes:\nx = (-b +,- SQRT(b^2 - 4ac))/2a\na = 3, b = -5, c = 2;\nx = (-(-5) +,- SQRT(25 - 24))/ 6\nx = (5 + 1) / 6; x = (5 - 1) / 6\nx = (1, 2/3)\nThis would have looked a lot better if I had my math editor!!!\n\nClarification: I solved your quadratic equation using the quadratic formula.\n\nMashi below simplified your quadratic equation, which you don't need the quadratic formula to do. Substitute the values of x I obtained by solving into the simplified equation, they will equal 0
146612
A quadratic equation is a second-order polynomial equation in a single variable  \n\n (1) \n\nwith . Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has two solutions. These solutions may be both real, or both complex. \n\nThe roots  can be found by completing the square, \n\n (2) \n\n (3) \n\n (4) \n\nSolving for  then gives \n\n (5) \n\nThis equation is known as the quadratic formula. \n\nThe first known solution of a quadratic equation is the one given in the Berlin papyrus from the Middle Kingdom (ca. 2160-1700 BC) in Egypt. This problem reduces to solving \n\n   (6) \n   (7) \n\n(Smith 1953, p. 443). The Greeks were able to solve the quadratic equation by geometric methods, and Euclid's  (ca. 325-270 BC) Data contains three problems involving quadratics. In his work Arithmetica, the Greek mathematician Diophantus  (ca. 210-290) solved the quadratic equation, but giving only one root, even when both roots were positive (Smith 1951, p. 134). \n\nA number of Indian mathematicians gave rules equivalent to the quadratic formula. It is possible that certain altar constructions dating from ca. 500 BC represent solutions of the equation, but even should this be the case, there is no record of the method of solution (Smith 1953, p. 444). The Hindu mathematician Aryabhata (475 or 476-550) gave a rule for the sum of a geometric series that shows knowledge of the quadratic equations with both solutions (Smith 1951, p. 159; Smith 1953, p. 444), while Brahmagupta  (ca. 628) appears to have considered only one of them (Smith 1951, p. 159; Smith 1953, pp. 444-445). Similarly, Mahavira (ca. 850) had substantially the modern rule for the positive root of a quadratic. Sridhara (ca. 1025) gave the positive root of the quadratic formula, as stated by Bhaskara (ca. 1150; Smith 1953, pp. 445-446). The Persian mathematicians al-Khwarizmi  (ca. 825) and Omar Khayyám  (ca. 1100) also gave rules for finding the positive root. \n\nViète  was among the first to replace geometric methods of solution with analytic ones, although he apparently did not grasp the idea of a general quadratic equation (Smith 1953, pp. 449-450). \n\nAn alternate form of the quadratic equation is given by dividing (◇) through by : \n\n (8) \n\n (9) \n\n (10) \n\nTherefore, \n\n (11) \n\n (12) \n\n (13) \n\nThis form is helpful if , where  denotes much greater, in which case the usual form of the quadratic formula can give inaccurate numerical results for one of the roots. This can be avoided by defining \n\n (14) \n\nso that  and the term under the square root sign always have the same sign. Now, if , then \n\n (15) \n\n   (16) \n   (17) \n\nso \n\n   (18) \n   (19) \n\nSimilarly, if , then \n\n (20) \n\n   (21) \n   (22) \n\nso \n\n   (23) \n   (24) \n\nTherefore, the roots are always given by  and . \n\nNow consider the equation expressed in the form \n\n (25) \n\nwith solutions  and . These solutions satisfy Vieta's formulas \n\n   (26) \n   (27) \n\nThe properties of the symmetric polynomials appearing in Vieta's formulas then give \n\n   (28) \n   (29) \n   (30) \n\n \nGiven a quadratic integer polynomial , consider the number such polynomials that are factorable over the integers for  and  taken from some set of integers . For example, for , there are four such polynomials, \n\n   (31) \n   (32) \n   (33) \n   (34) \n\nThe following table summarizes the counts of such factorable polynomials for simple  and small . Plots of the fractions of factorable polynomials for  (red),  (blue), and  (green) are also illustrated above. Amazingly, the sequence for  has the recurrence equation \n\n (35) \n\nwhere  is the number of divisors of  and  is the characteristic function of the square numbers. \n\n Sloane factorable over  for , 1, ... \n A067274 1, 4, 10, 16, 25, 31, 41, 47, 57, ... \n A091626 1, 2, 4, 6, 9, 11, 14, 16, 19, 22, ... \n A091627 0, 0, 1, 2, 4, 5, 7, 8, 10, 12, 14, ...
958565
Well, for quadratics, which are polynomials of degree 2 (i.e. the largest term is ax^2), the quadratic formula always works.  Of course, it's scary looking if square roots intimidate you.
839343
x^2 + 4x + 4 = 7 (subtrat 7 from both sides)\n\nx^2 + 4x -3 = 0 (use the quadratic equation formula)\n\n-b + or - the square root of (b^2 - 4ac) / a^2\n\n-4 + or - the square root of 16 - (4) (1) (-3) / 2\n\n-4 + or - the square root of 28 / 2\n\n-4 + or - (the square root of 4) (the square root of 7) / 2\n\n-4 + or - 2 (the square root of 7) / 2 (simplify by dividing 2 into  the -4 and 2 on the top to get rid of the 2 on the bottom)\n\nthat gives you x = - 2 + or - the square root of 7.
955903
3x^4-2x^2-8=0\nQuadratic formula with x^2 instead of x\n\na=3, b=-2, c=-8\n[2+-sqrt(4-4(3)(-8)]/6\n(2+-10)/6\nx^2=2 or x^2=-4/3\n\nx=+2 or -2\nx can also equal +sqrt(-4/3) or -sqrt(-4/3) but these are complex numbers
1072562
I'm reading the equation as \n\n4x^2 + 8x -1 = 0\n\nUsing quadratic formula x = (-8 +/- sqrt(64+16))/8 \n\nso x = -18/8 or 2/8 = -9/4 or 1/4\n\nThose are the roots.\n\nSubstitute each of those values as y into the second formula.
1122728
in standard ntation we have:\n\nA=2\nB=sqrt[3]\nC=1\n\ntan2t = B/(C-A) = -sqrt[3]\nHence 2t=-60 deg\nt=-30 deg\n\nThe standard notation for a quadratic curve is:\nAx^2+Bxy+Cy^2+Dx+Ey+F=0\nA, B, C determine the shape of the curve.\nThey are affected by rotation.\nchanging D, E, F only changes the position of the curve in plane by displacement.
569999
please rewrite this equation...the equal sign and/or the parenthesis seem to be misplaced.\n\nif you meant:\n[(x^3+27)/(9x+27)]/(3x^2-9x) = 27/(4x)  then:\n\ncross multiply and:\n\n(4x)*(x^3+27) = 27*[(9x+27)*(3x^2-9x)]\n(4x)*(x+3)*(x^2-3x+9) = 27*9*3x*[(x+3)*(x-3)]\n4(x^2-3x+9) = 729*(x-3)\n4x^2 - 12x -729x +36 + 2187 = 0\n4x^2 - 741x + 2223 = 0\n\nuse quadratic formula (-b +/- sqrt(b^2-4ac))/2a\n(x = 48.8035724297943  and  2915.19642757021)\n\n\nif you meant:\n[(x^3+27)*(9x+27)]/(3x^2-9x) = 27/(4x)  then:\n\ncross multiply and simplify\n\nif neither of these are the equation you were given, then repost using proper notation.