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

The solutions to the equation -2x^2-x=6 are x=-1/4-i(sqrt(47))/4 ,x=-1/4+i(sqrt(47))/4

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

The zeros of -2x^2-x=6 are x=-1/4-i(sqrt(47))/4 ,x=-1/4+i(sqrt(47))/4

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- 2 x^{2} - x = 6

x = - \frac{1}{4} - i \frac{47}{4}, x = - \frac{1}{4} + i \frac{47}{4}

Solution steps

- 2 x^{2} - x = 6

Move

6

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 (- 2) (- 6) : 47 i

x_{1, 2} = \frac{- ( - 1 ) \pm 47 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 47 i}{2 ( - 2 )}, x_{2} = \frac{- ( - 1 ) - 47 i}{2 ( - 2 )}

x = \frac{- ( - 1 ) + 47 i}{2 ( - 2 )} : - \frac{1}{4} - i \frac{47}{4}

x = \frac{- ( - 1 ) - 47 i}{2 ( - 2 )} : - \frac{1}{4} + i \frac{47}{4}

The solutions to the quadratic equation are:

x = - \frac{1}{4} - i \frac{47}{4}, x = - \frac{1}{4} + i \frac{47}{4}

y, U = X^{3} + 3 X y^{2}

\frac{x ^{2}}{2} = \frac{2 x}{4}

12 = 10 t + \frac{1}{2} (- 3.7) t^{2}

y, x^{2} y = 5

- 5 x^{2} + 9 x - 3 = 0

What are the solutions to the equation -2x^2-x=6 ?
The solutions to the equation -2x^2-x=6 are x=-1/4-i(sqrt(47))/4 ,x=-1/4+i(sqrt(47))/4
Find the zeros of -2x^2-x=6
The zeros of -2x^2-x=6 are x=-1/4-i(sqrt(47))/4 ,x=-1/4+i(sqrt(47))/4