What are the solutions to the equation 1=-2x^2-x ?

The solutions to the equation 1=-2x^2-x are x=-1/4-i(sqrt(7))/4 ,x=-1/4+i(sqrt(7))/4

Find the zeros of 1=-2x^2-x

The zeros of 1=-2x^2-x are x=-1/4-i(sqrt(7))/4 ,x=-1/4+i(sqrt(7))/4

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1 = - 2 x^{2} - x

x = - \frac{1}{4} - i \frac{7}{4}, x = - \frac{1}{4} + i \frac{7}{4}

Solution steps

1 = - 2 x^{2} - x

Switch sides

- 2 x^{2} - x = 1

Move

1

to the left side

- 2 x^{2} - x - 1 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 (- 2) (- 1) : 7 i

x_{1, 2} = \frac{- ( - 1 ) \pm 7 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 7 i}{2 ( - 2 )}, x_{2} = \frac{- ( - 1 ) - 7 i}{2 ( - 2 )}

x = \frac{- ( - 1 ) + 7 i}{2 ( - 2 )} : - \frac{1}{4} - i \frac{7}{4}

x = \frac{- ( - 1 ) - 7 i}{2 ( - 2 )} : - \frac{1}{4} + i \frac{7}{4}

The solutions to the quadratic equation are:

x = - \frac{1}{4} - i \frac{7}{4}, x = - \frac{1}{4} + i \frac{7}{4}

4 (2 - 3 x) > 5 (5 - 3 x)

4^{2} + 2^{2} = c^{2}

30 x - 18 > 252

\frac{2}{5} + \frac{5}{20}

- 4.3 x - 1.8 = 7.5 x - 6.3

What are the solutions to the equation 1=-2x^2-x ?
The solutions to the equation 1=-2x^2-x are x=-1/4-i(sqrt(7))/4 ,x=-1/4+i(sqrt(7))/4
Find the zeros of 1=-2x^2-x
The zeros of 1=-2x^2-x are x=-1/4-i(sqrt(7))/4 ,x=-1/4+i(sqrt(7))/4