Question

Find a number that is 56 less than its square.
help pls!

Answer 1

Let x be the number. Then, we have: x = x^2 - 56. This is just: \[ \Large {x^2 - x - 56 = 0} \]This can easily be factored to: (x + 7)(x - 8) = 0, so x = -7 or 8 <- two numbers that satisfy the given condition. Hope that helped! :)

Answer 2

x = x^2 - 56
x^2 - 56 - x = 0
x^2 - x - 56 = 0
x^2 -8x + 7x -7*8 = 0
x(x-8) + 7(x - 8) = 0
(x - 8)(x + 7) = 0

x - 8 = 0
x + 7 = 0

x = 8
x =-7

Answer 3

@Ahaanomegas & @Hero thanks so much for your help! :D