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

The solutions to the equation 2-x^2=x are x=-2,x=1

Find the zeros of 2-x^2=x

The zeros of 2-x^2=x are x=-2,x=1

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2 - x^{2} = x

x = - 2, x = 1

Solution steps

2 - x^{2} = x

Move

x

to the left side

2 - x^{2} - x = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 (- 1) \cdot 2 = 3

x_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 3}{2 ( - 1 )}, x_{2} = \frac{- ( - 1 ) - 3}{2 ( - 1 )}

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

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

The solutions to the quadratic equation are:

x = - 2, x = 1

x^{2} - 3 x

x^{2} + 8 x + 15 = 0

16 x^{2} + 16 x - 5 = 0

- 3 x^{2} + 3 = 0

x^{2} + 17 x + 60 = 0

What are the solutions to the equation 2-x^2=x ?
The solutions to the equation 2-x^2=x are x=-2,x=1
Find the zeros of 2-x^2=x
The zeros of 2-x^2=x are x=-2,x=1