What are the solutions to the equation -2x^2-3=8x+6 ?

The solutions to the equation -2x^2-3=8x+6 are x=-2-i(sqrt(2))/2 ,x=-2+i(sqrt(2))/2

Find the zeros of -2x^2-3=8x+6

The zeros of -2x^2-3=8x+6 are x=-2-i(sqrt(2))/2 ,x=-2+i(sqrt(2))/2

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

x = - 2 - i \frac{2}{2}, x = - 2 + i \frac{2}{2}

Solution steps

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

Move

6

to the left side

- 2 x^{2} - 9 = 8 x

Move

8 x

to the left side

- 2 x^{2} - 9 - 8 x = 0

Write in the standard form

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

- 2 x^{2} - 8 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 2 ) ( - 9 )}{2 ( - 2 )}

Simplify

(- 8)^{2} - 4 (- 2) (- 9) : 22 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 2 - i \frac{2}{2}, x = - 2 + i \frac{2}{2}

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

4 x^{2} + 9 x - 7 = 2

9 x^{2} - 6 x = 8

3 v^{2} + 63 v - 66 = 0

4 a^{2} + 4 a + 1 = 0

What are the solutions to the equation -2x^2-3=8x+6 ?
The solutions to the equation -2x^2-3=8x+6 are x=-2-i(sqrt(2))/2 ,x=-2+i(sqrt(2))/2
Find the zeros of -2x^2-3=8x+6
The zeros of -2x^2-3=8x+6 are x=-2-i(sqrt(2))/2 ,x=-2+i(sqrt(2))/2