What are the solutions to the equation -x^2+6=x^2+4x ?

The solutions to the equation -x^2+6=x^2+4x are x=-3,x=1

Find the zeros of -x^2+6=x^2+4x

The zeros of -x^2+6=x^2+4x are x=-3,x=1

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- x^{2} + 6 = x^{2} + 4 x

x = - 3, x = 1

Solution steps

- x^{2} + 6 = x^{2} + 4 x

Move

4 x

to the left side

- x^{2} + 6 - 4 x = x^{2}

Move

x^{2}

to the left side

- 2 x^{2} - 4 x + 6 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8}{2 ( - 2 )}, x_{2} = \frac{- ( - 4 ) - 8}{2 ( - 2 )}

x = \frac{- ( - 4 ) + 8}{2 ( - 2 )} : - 3

x = \frac{- ( - 4 ) - 8}{2 ( - 2 )} : 1

The solutions to the quadratic equation are:

x = - 3, x = 1

x^{2} + 6 x - 19 = 23

x^{2} - \frac{8}{3} x + \frac{16}{9} = 0

x^{2} + 2 x - 70 = 0

3 p^{2} - 3 p - 6 = - 2

64 x^{2} - 212 x + 49 = 0

What are the solutions to the equation -x^2+6=x^2+4x ?
The solutions to the equation -x^2+6=x^2+4x are x=-3,x=1
Find the zeros of -x^2+6=x^2+4x
The zeros of -x^2+6=x^2+4x are x=-3,x=1