What are the solutions to the equation x=-2x^2-12x-4 ?

The solutions to the equation x=-2x^2-12x-4 are x=-(13+sqrt(137))/4 ,x=-(13-sqrt(137))/4

Find the zeros of x=-2x^2-12x-4

The zeros of x=-2x^2-12x-4 are x=-(13+sqrt(137))/4 ,x=-(13-sqrt(137))/4

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x = - 2 x^{2} - 12 x - 4

x = - \frac{13 + 137}{4}, x = - \frac{13 - 137}{4}

Decimal Notation

x = - 6.17617 \dots, x = - 0.32382 \dots

Solution steps

x = - 2 x^{2} - 12 x - 4

Switch sides

- 2 x^{2} - 12 x - 4 = x

Move

x

to the left side

- 2 x^{2} - 13 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 ( - 2 ) ( - 4 )}{2 ( - 2 )}

(- 13)^{2} - 4 (- 2) (- 4) = 137

x_{1, 2} = \frac{- ( - 13 ) \pm 137}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 137}{2 ( - 2 )}, x_{2} = \frac{- ( - 13 ) - 137}{2 ( - 2 )}

x = \frac{- ( - 13 ) + 137}{2 ( - 2 )} : - \frac{13 + 137}{4}

x = \frac{- ( - 13 ) - 137}{2 ( - 2 )} : - \frac{13 - 137}{4}

The solutions to the quadratic equation are:

x = - \frac{13 + 137}{4}, x = - \frac{13 - 137}{4}

∣ 7 y ∣ = 0

k^{2} - 16 k = 17

x^{3} + x^{2} + 16 x + 16

x = 4!

m^{2} + 8 m

What are the solutions to the equation x=-2x^2-12x-4 ?
The solutions to the equation x=-2x^2-12x-4 are x=-(13+sqrt(137))/4 ,x=-(13-sqrt(137))/4
Find the zeros of x=-2x^2-12x-4
The zeros of x=-2x^2-12x-4 are x=-(13+sqrt(137))/4 ,x=-(13-sqrt(137))/4