What are the solutions to the equation 4-x^2=2+x ?

The solutions to the equation 4-x^2=2+x are x=-2,x=1

Find the zeros of 4-x^2=2+x

The zeros of 4-x^2=2+x are x=-2,x=1

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4 - x^{2} = 2 + x

x = - 2, x = 1

Solution steps

4 - x^{2} = 2 + x

Move

x

to the left side

4 - x^{2} - x = 2

Move

2

to the left side

- 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} + \frac{x}{2} = 5

6 x^{2} - 8 x = 3 x - 4

4 x^{2} + 3 x = 24 - x

5 t = t^{2}

\frac{x ^{2}}{3} - 192 = 0

What are the solutions to the equation 4-x^2=2+x ?
The solutions to the equation 4-x^2=2+x are x=-2,x=1
Find the zeros of 4-x^2=2+x
The zeros of 4-x^2=2+x are x=-2,x=1