What are the solutions to the equation -4x^2-19=x ?

The solutions to the equation -4x^2-19=x are x=-1/8-i(sqrt(303))/8 ,x=-1/8+i(sqrt(303))/8

Find the zeros of -4x^2-19=x

The zeros of -4x^2-19=x are x=-1/8-i(sqrt(303))/8 ,x=-1/8+i(sqrt(303))/8

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- 4 x^{2} - 19 = x

x = - \frac{1}{8} - i \frac{303}{8}, x = - \frac{1}{8} + i \frac{303}{8}

Solution steps

- 4 x^{2} - 19 = x

Move

x

to the left side

- 4 x^{2} - 19 - x = 0

Write in the standard form

a x^{2} + b x + c = 0

- 4 x^{2} - x - 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 4 ) ( - 19 )}{2 ( - 4 )}

Simplify

(- 1)^{2} - 4 (- 4) (- 19) : 303 i

x_{1, 2} = \frac{- ( - 1 ) \pm 303 i}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 303 i}{2 ( - 4 )}, x_{2} = \frac{- ( - 1 ) - 303 i}{2 ( - 4 )}

x = \frac{- ( - 1 ) + 303 i}{2 ( - 4 )} : - \frac{1}{8} - i \frac{303}{8}

x = \frac{- ( - 1 ) - 303 i}{2 ( - 4 )} : - \frac{1}{8} + i \frac{303}{8}

The solutions to the quadratic equation are:

x = - \frac{1}{8} - i \frac{303}{8}, x = - \frac{1}{8} + i \frac{303}{8}

5 x^{2} + 2 x = 3

x, 4 x^{2} - 4 m x + 5 m = 0

- 6 x^{2} + 8 x - 5 = 0

x^{2} - 47 x = 0

x^{2} + 3 x = x + 2

What are the solutions to the equation -4x^2-19=x ?
The solutions to the equation -4x^2-19=x are x=-1/8-i(sqrt(303))/8 ,x=-1/8+i(sqrt(303))/8
Find the zeros of -4x^2-19=x
The zeros of -4x^2-19=x are x=-1/8-i(sqrt(303))/8 ,x=-1/8+i(sqrt(303))/8