what is the quatratic formula?
482048
-b (+/-) sqrt(b^2 - 4ac)\n--------------------------------- = x\n2a\n\nUsing the equation "ax^2 + bx + c = 0" to find the values of a, b, and c.
641623
(-6)^2 - 4(9)(1) = 0\n\nThe discriminant is b^2 - 4ac, where a, b, and c are such that\n\nax^2 + bx + c = 0\n\nFor your formula, 9x^2 - 6x + 1 = 0, so a = 9, b = -6, and c = 1\n\nSince the discriminant is zero, there is only one real root to this equation.  If the discriminant is positive, then the quadratic has two real roots.  If the discriminant is negativ, then there are no real roots (both roots are complex).  This only has one root because\n\n9x^2-6x+1 = (3x - 1)^2 = 0.\n\nSo 3x-1 = 0.\nSo 3x = 1.\nSo x = 1/3.
1057648
You always can bring a quadratic equation\nax2 + bx + c = 0  in the form x2 + 2px + q = 0\nby dividing with a and p = b/(2a) while q = c/a\n\nx2 + 2px + q = 0\nx2 + 2px + p2 + q = p2\n(x + p)^2 = p2 - q\nx + p = +V(p2 - q) or x + p =  -V(p2 - q) \nx  = -p +V(p2 - q) or x = -p  - V(p2 - q)\n\nWhen you replace p and q you find the famous abc formula.\nSatisfied?
1021342
quadratic formula would probably be easiest since all you need is a calculator.
597401
a^2 + 321 a - 51 = ( a - 0.158799954)*(a + 321.158799954)\nUsing the general law for solving a quadratic equation.
953700
The discriminant\n\n√b² - 4ac portion of the quadratic Formula\n\nx = - b ± √b² - 4ac/ 2a is negative , no real number solutions \n\n- - - - - -s-
217158
It's abc formula to solve the quadratic equation but the plus should be plus or minus
962761
treat it as a quadratic equation and use the quadratic formula:\n\nfirst subtract A from both sides:\n2pr^2 +2prh - A = 0\n\nthen plug the coefficients: 2p, 2ph, -A into the\nquadratic formula for a, b, c, respectively and simplify:\n\nr = [- b +/- sqrt(b^2 - 4ac)]/(2a)\n\ni'll leave it to you to do the tedious steps to get\nr = [ph +/- sqrt(p^2h^2 + 2pA)]/(2p)
766061
p^9-q^9/p^6-q^6\n=(p^3-q^3)(p^6+p^3q^3+q^6)/(p^3+q^3)(p^3-q^3)\n=(p^6+p^3q^3+q^6)/(p^3+q^3)
705433
Derivation of the quadratic formula by completing the square.\n\nax^2 + bx + c = 0\nx^2 + (b/a)x + (c/a) = 0\nx^2 + (b/a)x = -c/a\nx^2 + (b/a)x + b^2/(4a^2) =  b^2/(4a^2) - c/a \n(x + b/2a)^2 =  b^2/(4a^2) - 4ac/(4a^2) \nx + b/2a = ±√[ b^2/(4a^2) - 4ac/(4a^2) ]\n\nx = -b/2a ± √[ b^2/(4a^2) - 4ac/(4a^2) ]\n= -b/2a ± √[ b^2- 4ac] / 2a\n\nx = {-b ± √[ b^2- 4ac] } / 2a
443949
Use quadratic formula and do your own homework.
508029
The equation is:\n(x+a+b+c)^2 = 2x^2+(x+a)^2-(x+b)^2-(x+c)^2\nA simple rearrangement yields\n(x+a+b+c)^2 - x^2 = (x+a)^2-(x+b)^2 + x^2-(x+c)^2\nFactorize using the identity a^2-b^2=(a+b)(a-b)\n(x+a+b+c+x)(x+a+b+c-x) = (x+a+x+b)(x+a-x-b) + (x+x+c)(x-x-c)\nwhich can be written as\n(2x+a+b+c)(a+b+c) = (2x+a+b)(a-b)-c(2x+c)\n==> 2x(a+b+c)+(a+b+c)^2 = 2x(a-b)+(a-b)(a+b)-2xc-c^2\n==> 2x(a+b+c-a+b+c) = (a+b+c)^2 + a^2-b^2-c^2\n==> 4x(b+c) = a^2+b^2+c^2+2ab+2bc+2ca+a^2-b^2-c^2\n            = 2(a^2+ab+bc+ca) = 2(a+b)(c+a)\n==> x = (a+b)(c+a)/[2*(b+c)]
797913
That's the quadratic formula. It is the solution to equations of the form\nax^2+bx+c=0
688944
View source of the that page.\nYou will see the formulas.
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, ...
924128
quark's  formula is slight incorrect\n\nVescape = sqrt(2GM/r)\n\nwhere M is mass of the body being escaped from (i.e the planet or star or asteroid)\n\nand r is the distance between the object and the body being\nescaped from\n\nand G is the gravitational constant.\n\nTo get this formula is a engery equation\n\ninitial kinetic energy required is the potential energy needed\nto bring the potential energy from current state to state\nat inifinite distance (i.e to zero)\n\nIn a gravitational field the potentail engery = - GMm/r\nfor object of mass m.\n\nSo to get this to zero you need a kinetic energy of\n\n1/2mv^2 = GMm/r\n\nsolve for v and you get sqrt(2GM/r).
89652
The formula involves complex maths calculations. Ask some maths professor.