What are the solutions to the equation -x^2+12=x ?

The solutions to the equation -x^2+12=x are x=-4,x=3

Find the zeros of -x^2+12=x

The zeros of -x^2+12=x are x=-4,x=3

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- x^{2} + 12 = x

x = - 4, x = 3

Solution steps

- x^{2} + 12 = x

Move

x

to the left side

- x^{2} + 12 - x = 0

Write in the standard form

a x^{2} + b x + c = 0

- x^{2} - x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 12}{2 ( - 1 )}

(- 1)^{2} - 4 (- 1) \cdot 12 = 7

x_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 7}{2 ( - 1 )}, x_{2} = \frac{- ( - 1 ) - 7}{2 ( - 1 )}

x = \frac{- ( - 1 ) + 7}{2 ( - 1 )} : - 4

x = \frac{- ( - 1 ) - 7}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

x = - 4, x = 3

3 x^{2} - 2 x - 2 = - 1

x^{2} + (n + 1) x + n = 0

5 x^{2} + 9 x + 1

x^{2} - 8 x + 12 = 0

a, y = a^{2}

What are the solutions to the equation -x^2+12=x ?
The solutions to the equation -x^2+12=x are x=-4,x=3
Find the zeros of -x^2+12=x
The zeros of -x^2+12=x are x=-4,x=3