What are the solutions to the equation 8k^2+12=-2k ?

The solutions to the equation 8k^2+12=-2k are k=-1/8+i(sqrt(95))/8 ,k=-1/8-i(sqrt(95))/8

Find the zeros of 8k^2+12=-2k

The zeros of 8k^2+12=-2k are k=-1/8+i(sqrt(95))/8 ,k=-1/8-i(sqrt(95))/8

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8 k^{2} + 12 = - 2 k

k = - \frac{1}{8} + i \frac{95}{8}, k = - \frac{1}{8} - i \frac{95}{8}

Solution steps

8 k^{2} + 12 = - 2 k

Move

2 k

to the left side

8 k^{2} + 12 + 2 k = 0

Write in the standard form

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

8 k^{2} + 2 k + 12 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 8 \cdot 12}{2 \cdot 8}

Simplify

2^{2} - 4 \cdot 8 \cdot 12 : 295 i

k_{1, 2} = \frac{- 2 \pm 2 95 i}{2 \cdot 8}

Separate the solutions

k_{1} = \frac{- 2 + 2 95 i}{2 \cdot 8}, k_{2} = \frac{- 2 - 2 95 i}{2 \cdot 8}

k = \frac{- 2 + 2 95 i}{2 \cdot 8} : - \frac{1}{8} + i \frac{95}{8}

k = \frac{- 2 - 2 95 i}{2 \cdot 8} : - \frac{1}{8} - i \frac{95}{8}

The solutions to the quadratic equation are:

k = - \frac{1}{8} + i \frac{95}{8}, k = - \frac{1}{8} - i \frac{95}{8}

x, f (x^{2}) = 3 x + 2

y^{2} - 2 y - 9 = 0

x^{2} = 1^{2} + 3^{2}

(M^{2} - M - 5) = 0

y (y - 3) = 4

What are the solutions to the equation 8k^2+12=-2k ?
The solutions to the equation 8k^2+12=-2k are k=-1/8+i(sqrt(95))/8 ,k=-1/8-i(sqrt(95))/8
Find the zeros of 8k^2+12=-2k
The zeros of 8k^2+12=-2k are k=-1/8+i(sqrt(95))/8 ,k=-1/8-i(sqrt(95))/8