if x2 + 7 = 37, then x2 = 29 and x = ±√29
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∙ 9y agox2+6x-7 = 0 (x+3)2-7 = 0 (x+3)2-7-9 = 0 (x+3)2 = 16 x+3 = + or - 4 x = - 3 + or - 4 x = 1 or -7
C and D
2x2+7/x1
2x = -7 - 8y
x2+7x x(x+7)
x2+12x+36=0 2x+12x+36=0 14x+36=0 14x=-36 x=-36/14 x=-2 4/7
x2 + 10x + 21 = (x + 3)(x + 7)
7
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
x2 + 3x + 7 = 5 ∴ x2 + 3x + 2 = 0 ∴ (x + 1)(x + 2) = 0 ∴ x ∈ {-2, -1}
x=-0.7
7
A quadratic equation. If you wish to solve for x, you can do so as follows: -x2 + 6x + 7 = 0 x2 - 6x - 7 = 0 (x - 7)(x + 1) = 0 x ∈ {-1, 7}
(x+ 7)(x + 1) so x = -1 or -7
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
x2 + 3x - 7 = 5x + 8; thus, x2 + 3x - 7 - 5x - 8 = x2 - 2x - 15 = (x - 5)(x + 3) = 0. Therefore the solution is x = 5 or -3.
The answer is (3.5)2 plus (4)2 equals 410.