how would you solve by completing the square and using the quadratic formula?
342699
Completing the square involves creating a perfect square trinomial.  Consider\n\nx^2 + 4x - 1 = 0\n\nCompleting the square would involve trying to make one side a perfect square: (x + 2)^2 in this case.  Step by step:\n\n1.  Get all your x terms on one side, and constants on the other.\n\nx^2 + 4x = 1\n\n2.  Take half of the coefficient of the x term and square it.  Half of 4 is 2, and 2 squared is 4.\n\n3.  Add this to both sides of the equation.\n\nx^2 + 4x +4 = 1 + 4\n\n4.  Now factor the left side.  It works out to be a perfect square.\n\n(x + 2)^2 = 5\n\n5.  Take the square root of both sides and solve for x.\n\nx + 2 = +/- sqrt(5)\n\nx = +/- sqrt(5) - 2\n\nThe quadratic equation is pretty easy:\n\nx = [-b +/- sqrt (b^2 - 4ac)] / 2a\n\nIn the above example, a = 1, b = 4, and c = -1 (coefficients of the x^2 and x terms are a and b, and the constant is c).  So plugging in to the quadratic formula gives you\n\nx = [-4 +/- sqrt(4^2 - 4x1x(-1))] / 2x1\n  = [-4 +/- sqrt(16 - (-4))] / 2\n  = [-4 +/- sqrt(20)] / 2\n  = [-4 +/- 2sqrt(5)] / 2\n  = -2 +/- sqrt(5)\n\nwhich is the same as we got via completing the square.
797913
That's the quadratic formula. It is the solution to equations of the form\nax^2+bx+c=0
1021342
quadratic formula would probably be easiest since all you need is a calculator.
781653
Eliminate the y to get\n2-x^2=2x+4\nx^2+2x+2=0\nusing quadratic formula\n(-2+-sqrt(4-4*2))/2\nwe see there are no real roots: the two functions do not intersect\nx= -1+-i
734091
This particular quadratic equation will require the quadratic formula:\n\n-b +/- sqrt(b^2-4ac) all divided by 2a.\n\nHere, a=1, b=18, and c=49\n\n-18 +/- sqrt[(18^2 - 4(1)(49)] all divided by 2.\n\n18^2 - 4 x 49 = 128.....\n\nLet's do the top first:\n\n-18 +/- sqrt(128)\n\nWhat is sqrt(128)? You need to factor 128 = 64 x 2\nTherefore, sqrt(128) = sqrt(64 x 2) = 8 x sqrt(2).\n(we took 64 out of the root function, so comes out as the sqrt, which is 8.)\n\nSet it all up:\n\n-18 +/- 8sqrt(2) divided by 2:\n\n-9 +/- 4sqrt(2)\n\nAnswer: -9 - 4sqrt(2), -9 + 4sqrt(2).\n\nNot easy due to the answer not a set of integers.
406680
Do you mean 6x^2-3x-4?\n\nIf you do, we know that there are 3 ways of solving quadratics: factoring, completing the square, and quadratic formula. I would not recommend completing the square as it is long and involves tedious math word.\n\nAnother helpful hint: In standard form ax^2+bx+c=0, b^2-4ac is known as the discriminant. If it is negative, there are no solutions, and if it is not a perfect square, then you cannot factor.\n\nIn this quadratic, the discriminant would be (-3)^2-4(6)(-4) which equals 105. 105 is not a perfect square, so we cannot factor.\n\nQuadratic Formula: \n           -b +/-  sqrt (b^2-4ac)\nx= -------------------------------\n                2a\n\nsubstitute for a, b, and c
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
932849
The quadratic formula goes: (-b ± √b² - 4ac)/ 2a\nSo x²=a, -6x=b, -7=c\nNow plug it in:\n [-(-6) ± √(6)² -4(1)(-7)]/ 2(1) \nsimplify: (6± √36 + 28)/ 2\nsimplify further: (6±√64)/2\nwhich equals: (6±8)/2\nand if you separate it out: (6-8)/2 = -1      and   (6+8)/2 = 7
512145
subtract c from both sides then you'll have\n\nx^2 + Bx = -C\n\nthen you can complete the square\n\nx^2 +Bx +_____=-C +_______\n\nfill in the blank with (b/2)^2\n\ni'm positive this is right...i teach high school math.
595518
1 + 4.9t^2 = 6t\n4.9t^2 - 6t + 1 = 0\n\nusing the quadratic formula\n\nt = (-b ± sqrt(b^2 - 4ac))/2a\nt = (-(-6) ± sqrt((-6)^2 - 4(4.9)(1)))/2(4.9)\nt = (6 ± sqrt(36 - 19.6))/9.8\nt = (6 ± sqrt(16.4))/9.8\nt = (6 ± sqrt(4 * 4.1))/9.8\nt = (6 ± 2sqrt(4.1))/9.8\nt = (3 ± sqrt(4.1))/4.9\nt = about 1.02548 or .199011\n\n-----------------------------------\n\n1 - 4.9t^2 = 6t\n4.9t^2 + 6t - 1 = 0\n\nt = (-6 ± sqrt(36 + 4(4.9)))/9.8\nt = (-6 ± sqrt(36 + 16.4))/9.8\nt = (-6 ± sqrt(52.4))/9.8\nt = (-6 ± sqrt(4 * 13.1))/9.8\nt = (-6 ± 2sqrt(13.1))/9.8\nt = (-3 ± sqrt(13.1))/4.9\nt = about -1.37312 or .148627
443949
Use quadratic formula and do your own homework.
802559
You are given the quadratic function\ny = x² - 8x + 22\n\nTo find the vertex, change the given equation into the form\ny = a(x - h)² + k\n\nWhere \na = is a constant\n(h,k) is the vertex\n\nThus,\ny = x² - 8x + 22\n\nWe complete the square, by squaring the half of the coefficient of x (which is -8, the half of -8 is -4, and the square of -4 is 16.)  Thus,\ny = x² - 8x + 16 + 6\n\nNow, we can factor\ny = (x - 4)² + 6\n\nTherefore, the vertex is\n(4,6)\n\n^_^
239087
We can use the quadratic formula :\n\nx = -6 +/- sqrt(36 - 4(3)(-3))\n-----------------------------\n2(3)\n\nx = -6 +/- sqrt(36 + 36)\n-----------------------\n6\n\nx = -6 +/- sqrt(72)\n-------------\n6\n\nx = -6 +/- 6sqrt(2)\n-----------------\n6\n\nx = -1 +/- sqrt(2)\n\nANSWER : x = -1 + sqrt(2) or\nx = -1 - sqrt(2)
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
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)
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.
745544
the quadratic formula is a useful tool in solving an equation involving a variable and its square without breaking it down into smaller pieces. It can only be used when it is in the\nax^2+bx+c=0  format. (a, b, and c can be negative.)\n\nIn this problem, a=2, b=-2, and c=-1. The equation for the quadratic is:\n\nx= (-b+/- sq.rt (b^2-4ac))/2a\n\nSolving for the equation given, you get:\n\nx= (2 +/- sq.rt.(4-4*2*-1))/4\n   = (2+/- sq.rt 12)/4\n   = 1.366, 0.366