Assuming that by expressing that equation as a question you're looking to solve for x:
x2 - 4x - 29 = -10
x2 - 4x + 4 = 23
(x - 2)2 = 23
x - 2 = ± √23
x = 2 ± √23
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Use the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2aUse the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2aUse the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2aUse the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2a
if x2 + 7 = 37, then x2 = 29 and x = ±√29
There are 2 roots to the equation x2-4x-32 equals 0; factored it is (x-8)(x+4); therefore the roots are 8 & -4.
x2 = 11x - 10 ∴ x2 - 11x + 10 = 0 ∴ (x - 10)(x - 1) = 0 ∴ x ∈ {1, 10}
To solve for x: x2 = 11x - 10 x2 - 11x = -10 x2 - 11x + 10 =0 (x - 1)(x - 10) = 0 x = {1, 10)