What are the solutions to the equation 2n(4n+5)+6=0 ?

The solutions to the equation 2n(4n+5)+6=0 are n=-5/8+i(sqrt(23))/8 ,n=-5/8-i(sqrt(23))/8

Find the zeros of 2n(4n+5)+6=0

The zeros of 2n(4n+5)+6=0 are n=-5/8+i(sqrt(23))/8 ,n=-5/8-i(sqrt(23))/8

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2 n (4 n + 5) + 6 = 0

n = - \frac{5}{8} + i \frac{23}{8}, n = - \frac{5}{8} - i \frac{23}{8}

Solution steps

2 n (4 n + 5) + 6 = 0

Expand

2 n (4 n + 5) + 6 : 8 n^{2} + 10 n + 6

8 n^{2} + 10 n + 6 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 8 \cdot 6}{2 \cdot 8}

Simplify

1 0^{2} - 4 \cdot 8 \cdot 6 : 223 i

n_{1, 2} = \frac{- 10 \pm 2 23 i}{2 \cdot 8}

Separate the solutions

n_{1} = \frac{- 10 + 2 23 i}{2 \cdot 8}, n_{2} = \frac{- 10 - 2 23 i}{2 \cdot 8}

n = \frac{- 10 + 2 23 i}{2 \cdot 8} : - \frac{5}{8} + i \frac{23}{8}

n = \frac{- 10 - 2 23 i}{2 \cdot 8} : - \frac{5}{8} - i \frac{23}{8}

The solutions to the quadratic equation are:

n = - \frac{5}{8} + i \frac{23}{8}, n = - \frac{5}{8} - i \frac{23}{8}

- z^{2} + 3 z - 5 = 0

512

\frac{2}{3} \cdot \frac{2}{5}

c^{2} - 10 c + 25

\frac{3}{4} (x - 1) + x = 1

What are the solutions to the equation 2n(4n+5)+6=0 ?
The solutions to the equation 2n(4n+5)+6=0 are n=-5/8+i(sqrt(23))/8 ,n=-5/8-i(sqrt(23))/8
Find the zeros of 2n(4n+5)+6=0
The zeros of 2n(4n+5)+6=0 are n=-5/8+i(sqrt(23))/8 ,n=-5/8-i(sqrt(23))/8