Solve for x.  x^2 = 5x + 2?
952526
x^2-5x-2=0\n\nuse the quadratic formula\n\n[5  +/- sqrt(25-4*1*-2)]/2\n\n[5  +/-   sqrt(25+ 8)]/2\n\nI distributed the 1/2...easier to type\n\n5/2  +/-sqrt33/2
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
982146
Just add and substract terms in both sides.\n\nSince \ny-x^2+8x\nis equal to 7, \nyou can add x^2 and substract 8x to both sides.\n(y-x^2+8x) +x^2-8x = y\n7 +x^2-8x = x^2-8x+7\n\nso y=x^2-8x+7\n\nby the way, if you factorize in terms of x then you'd get\ny=(x-1)(x-7)
764068
2x - 2 = x^2 - 4x + 3\nSet one side equal to 0\n0 = x^2 - 6x + 5\nFactor\n0 = (x - 5)(x - 1)\nSet each factor equal to 0\n0 = x - 5              0 = x - 1\nSolve\n5 = x                  1 = x
946842
1.solve by factoring:\nx^2 -5x +x-5 =0\nx(x-5)+1(x-5)=0\n(x+1)(x-5)=0\n=> either x-5=0 => x=5\nor x+1 =0 => x=-1\n\n2. by using qudrt formula:\nfor ax^2+bx+c=0\nx = [-b+-sqrt[b^2 -4ac]]/2a\nhere a=1, b=-4, c=-5\ntherefore x= [4+-sqrt[16+20]]/2\nx=[4+-6]/2\n=>\nx=[4+6]/2=5 and x=[4-6]/2=-1
57070
Starting with\nx^2 - 5x - 6.25 = 0\n\nMOVE the lone term to the right side of the equation.\n\nSo\n x^2 - 5x = 6.25\n\nTo comlete the square on the left:\nDIVIDE the coefficient of the x term (5) by two (2.5).  Then SQUARE it.\n\nADD \n 2.5^2 (which happens to be 6.25) to both sides\n\nNow\n x^2 - 5x + 6.25 = 6.25 + 6.25\n\nREWRITE the left which is now a perfect square\n (x-2.5)^2 = 12.5 (or 25/2)\n\nTAKE THE SQUARE ROOT OF BOTH SIDES\n x - 2.5 = +or- Square root of 12.5\n\nANSWER\n x = Sqrt 12.5 + 2.5   -or-\n x = 2.5 - Sqrt 12.5
774115
.x^2+5x+6/x+1=6\n\nassuming you mean:\n(x^2+5x+6) / (x+1)=6\n\nx^2+5x+6=6 (x+1)\nx^2 -x=0\nx(x-1)=0\nso x=0 or x=1
657080
(x + 1)^2 = 25\nx + 1 = -5 or 5\nx = -6 or 4\n\n---------------------\n\nx^2 + 4x + 4 = 16\nx^2 + 4x - 12 = 0\n(x + 6)(x - 2) = 0\nx = -6 or 2\n\n----------------------\n\nx^2 - 14x + 49 = 13\nx^2 - 14x + 36 = 0\n\nx = (-b ± sqrt(b^2 - 4ac))/2a\nx = (-(-14) ± sqrt((-14)^2 - 4(1)(36)))/2(1)\nx = (14 ± sqrt(196 - 144))/2\nx = (14 ± sqrt(52))/2\nx = (14 ± sqrt(4 * 13))/2\nx = (14 ± 2sqrt(13))/2\nx = 7 ± sqrt(13)
1056888
2(x^2-4x)-1=\n2(x^2-4x+4) -8-1=\n2(x-2)^2-9
838051
5(x – 2)^2 = 3\n(x – 2)^2 =3/5\nx -2 =+- sqrt(3/5)\nx=2+- sqrt(3/5)
996406
solve for x for the eq. of x+ky=10, and you would get x=-ky+10. plug that eq. to the x^2+y^2=10. and solve for x or k. after you get that, plug it back into the x+ky=10.
98037
2. Take the square root of both sides, thereby negating the ^2 and giving 10. Then solve for 5x=10.
771896
expand x^2-6x+9 = 5\nor x^2-6x+4 =0
798535
x^3 - x^2 + 5x + 5\n\n(x^3 - x^2) + (5x - 5)\n\nx^2(x - 1) + 5(x - 1)\n\nsince both x^2 and 5 have the same factor, you can factor it to\n\n(x^2 + 5)(x - 1)\n\nif you want, you can either leave it that way, or\n\n(x + sqrt(5i))(x - sqrt(5i))(x - 1)\n\nwhat you can probably just leave it as (x^2 + 5)(x - 1)\n\nyou can also try www.quickmath.com, click Factor under Algebra, then type in x^3 - x^2 + 5x - 5, then click enter. \n\nAnymore Questions Ask ME
330086
Completing the square:\n\nGet the x^2 term by itself\nx^2 + 7x/5 - 6/5 = 0\n\nMove the constant to the other side\nx^2 + 7x/5 = 6/5\n\nTake half of the constant for the x term and square it: half of 7/5 is 7/10, squared is 49/100.  Add this to both sides of the equation\n\nx^2 + 7x/5 + 49/100 = 6/5 + 49/100\nx^2 + 7x/5 + 49/100 = 169/100\n\nThe left side is a perfect square trinomial - factor it\n(x + 7/10)^2 = 169/100\n\nTake the square root of both sides and solve for x\nx + 7/10 = +/- 13/10\nx = -7/10 +/- 13/10\nx = 6/10 or -20/10\nx = 0.6 or -2
783870
7x^2+16x over (x-2)(x+2)(x+3)
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1