Strictly speaking, it is a quadratic equation in x. x2 + x = 6 => x2 + x - 6 = 0 => (x+3)*(x-2) = 0 => x = -3 or x = 2
x2 + x - 30 = (x + 6)(x - 5)
x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1
-4
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
(x + 6)(x + 1) = 0 so x = either -1 or -6
x2 + x - 30 = (x + 6)(x - 5)
x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1
x2 + 7x + 9 = 3 ∴ x2 + 7x + 6 = 0 ∴ (x + 6)(x + 1) = 0 ∴ x ∈ {-6, -1}
-4
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
(x + 6)(x + 1) = 0 so x = either -1 or -6
Given that 52 - x2 + x = 10, then -52 + x2 - x + 10 = x2 - x - 42 = (x - 7)(x + 6) = 0. Therefore, x = 7 or -6.
x2 + 6x + 9 = 81 x2 + 6x = 72 x2 + 6x - 72 = 0 (x+12)(x-6) = 0 x= -12, 6 (two solutions)
x2- 7x + 6 = 0 factor, (x-6)(x-1) = 0 x = 6 x = 1
(x + 12)(x - 6) = 0 x = 6 or -12
x2 + X -6 =0 so X2 + X = 6 so X2 = 6-x so X = square root of (6-x) That's as far as I can get it.
x2+x-6=0 * * * * * Perhaps the solution you are looking for is: x2 + x - 6 = (x - 2)(x + 3) = 0; whence, x = 2 or -3.