What are the solutions to the equation -8n^2-3n+2=8 ?

The solutions to the equation -8n^2-3n+2=8 are n=-3/16-i(sqrt(183))/(16),n=-3/16+i(sqrt(183))/(16)

Find the zeros of -8n^2-3n+2=8

The zeros of -8n^2-3n+2=8 are n=-3/16-i(sqrt(183))/(16),n=-3/16+i(sqrt(183))/(16)

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- 8 n^{2} - 3 n + 2 = 8

n = - \frac{3}{16} - i \frac{183}{16}, n = - \frac{3}{16} + i \frac{183}{16}

Solution steps

- 8 n^{2} - 3 n + 2 = 8

Move

8

to the left side

- 8 n^{2} - 3 n - 6 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 8 ) ( - 6 )}{2 ( - 8 )}

Simplify

(- 3)^{2} - 4 (- 8) (- 6) : 183 i

n_{1, 2} = \frac{- ( - 3 ) \pm 183 i}{2 ( - 8 )}

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 183 i}{2 ( - 8 )}, n_{2} = \frac{- ( - 3 ) - 183 i}{2 ( - 8 )}

n = \frac{- ( - 3 ) + 183 i}{2 ( - 8 )} : - \frac{3}{16} - i \frac{183}{16}

n = \frac{- ( - 3 ) - 183 i}{2 ( - 8 )} : - \frac{3}{16} + i \frac{183}{16}

The solutions to the quadratic equation are:

n = - \frac{3}{16} - i \frac{183}{16}, n = - \frac{3}{16} + i \frac{183}{16}

\frac{x}{- 5} \geq 7

11 - 8

221 = 17 x

\frac{x - 1}{x + 1} > 0

5245

What are the solutions to the equation -8n^2-3n+2=8 ?
The solutions to the equation -8n^2-3n+2=8 are n=-3/16-i(sqrt(183))/(16),n=-3/16+i(sqrt(183))/(16)
Find the zeros of -8n^2-3n+2=8
The zeros of -8n^2-3n+2=8 are n=-3/16-i(sqrt(183))/(16),n=-3/16+i(sqrt(183))/(16)