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

The solutions to the equation 8=-x^2-x are x=-1/2-i(sqrt(31))/2 ,x=-1/2+i(sqrt(31))/2

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

The zeros of 8=-x^2-x are x=-1/2-i(sqrt(31))/2 ,x=-1/2+i(sqrt(31))/2

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8 = - x^{2} - x

x = - \frac{1}{2} - i \frac{31}{2}, x = - \frac{1}{2} + i \frac{31}{2}

Solution steps

8 = - x^{2} - x

Switch sides

- x^{2} - x = 8

Move

8

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 (- 1) (- 8) : 31 i

x_{1, 2} = \frac{- ( - 1 ) \pm 31 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 31 i}{2 ( - 1 )}, x_{2} = \frac{- ( - 1 ) - 31 i}{2 ( - 1 )}

x = \frac{- ( - 1 ) + 31 i}{2 ( - 1 )} : - \frac{1}{2} - i \frac{31}{2}

x = \frac{- ( - 1 ) - 31 i}{2 ( - 1 )} : - \frac{1}{2} + i \frac{31}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} - i \frac{31}{2}, x = - \frac{1}{2} + i \frac{31}{2}

\frac{3 x}{x + 4} \geq 0

e^{(\frac{0.9}{8.62 \cdot 1 0 ^{- 5} \cdot 1273})}

\frac{3}{4} - \frac{3}{8} \cdot \frac{1}{2}

5 (x + 1) = 4 (x + 8)

3 r^{2} + 18 r = 0

What are the solutions to the equation 8=-x^2-x ?
The solutions to the equation 8=-x^2-x are x=-1/2-i(sqrt(31))/2 ,x=-1/2+i(sqrt(31))/2
Find the zeros of 8=-x^2-x
The zeros of 8=-x^2-x are x=-1/2-i(sqrt(31))/2 ,x=-1/2+i(sqrt(31))/2