What are the solutions to the equation 4x-x^2+8=x^2-2x ?

The solutions to the equation 4x-x^2+8=x^2-2x are x=-1,x=4

Find the zeros of 4x-x^2+8=x^2-2x

The zeros of 4x-x^2+8=x^2-2x are x=-1,x=4

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4 x - x^{2} + 8 = x^{2} - 2 x

x = - 1, x = 4

Solution steps

4 x - x^{2} + 8 = x^{2} - 2 x

Move

2 x

to the left side

- x^{2} + 6 x + 8 = x^{2}

Move

x^{2}

to the left side

- 2 x^{2} + 6 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 2 ) \cdot 8}{2 ( - 2 )}

6^{2} - 4 (- 2) \cdot 8 = 10

x_{1, 2} = \frac{- 6 \pm 10}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 6 + 10}{2 ( - 2 )}, x_{2} = \frac{- 6 - 10}{2 ( - 2 )}

x = \frac{- 6 + 10}{2 ( - 2 )} : - 1

x = \frac{- 6 - 10}{2 ( - 2 )} : 4

The solutions to the quadratic equation are:

x = - 1, x = 4

- 2 x^{2} - 28 x + 30 = 0

(5 x^{2} + x)^{2} = (3 x + 3)^{2}

x, x^{2} + y^{2} - 6 x - 8 y = - 25

x - x^{2}

(x - 1) (x - 3) = - 1

What are the solutions to the equation 4x-x^2+8=x^2-2x ?
The solutions to the equation 4x-x^2+8=x^2-2x are x=-1,x=4
Find the zeros of 4x-x^2+8=x^2-2x
The zeros of 4x-x^2+8=x^2-2x are x=-1,x=4