What are the solutions to the equation 9=-4x^2-x ?

The solutions to the equation 9=-4x^2-x are x=-1/8-i(sqrt(143))/8 ,x=-1/8+i(sqrt(143))/8

Find the zeros of 9=-4x^2-x

The zeros of 9=-4x^2-x are x=-1/8-i(sqrt(143))/8 ,x=-1/8+i(sqrt(143))/8

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9 = - 4 x^{2} - x

x = - \frac{1}{8} - i \frac{143}{8}, x = - \frac{1}{8} + i \frac{143}{8}

Solution steps

9 = - 4 x^{2} - x

Switch sides

- 4 x^{2} - x = 9

Move

9

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 (- 4) (- 9) : 143 i

x_{1, 2} = \frac{- ( - 1 ) \pm 143 i}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 143 i}{2 ( - 4 )}, x_{2} = \frac{- ( - 1 ) - 143 i}{2 ( - 4 )}

x = \frac{- ( - 1 ) + 143 i}{2 ( - 4 )} : - \frac{1}{8} - i \frac{143}{8}

x = \frac{- ( - 1 ) - 143 i}{2 ( - 4 )} : - \frac{1}{8} + i \frac{143}{8}

The solutions to the quadratic equation are:

x = - \frac{1}{8} - i \frac{143}{8}, x = - \frac{1}{8} + i \frac{143}{8}

- a^{2} - 2 a + 3 = 0

a^{2} + 10 a - 12 = 0

a = a^{2} + (2 - \frac{4}{3} a)^{2} (\frac{3}{4})

(5 - x) (- 2 - x) + 12 = 0

5 x^{2} + 12 = 0

What are the solutions to the equation 9=-4x^2-x ?
The solutions to the equation 9=-4x^2-x are x=-1/8-i(sqrt(143))/8 ,x=-1/8+i(sqrt(143))/8
Find the zeros of 9=-4x^2-x
The zeros of 9=-4x^2-x are x=-1/8-i(sqrt(143))/8 ,x=-1/8+i(sqrt(143))/8