What are the solutions to the equation 6u^2+8u=-3 ?

The solutions to the equation 6u^2+8u=-3 are u=-2/3+i(sqrt(2))/6 ,u=-2/3-i(sqrt(2))/6

Find the zeros of 6u^2+8u=-3

The zeros of 6u^2+8u=-3 are u=-2/3+i(sqrt(2))/6 ,u=-2/3-i(sqrt(2))/6

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6 u^{2} + 8 u = - 3

u = - \frac{2}{3} + i \frac{2}{6}, u = - \frac{2}{3} - i \frac{2}{6}

Solution steps

6 u^{2} + 8 u = - 3

Move

3

to the left side

6 u^{2} + 8 u + 3 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

Simplify

8^{2} - 4 \cdot 6 \cdot 3 : 22 i

u_{1, 2} = \frac{- 8 \pm 2 2 i}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 8 + 2 2 i}{2 \cdot 6}, u_{2} = \frac{- 8 - 2 2 i}{2 \cdot 6}

u = \frac{- 8 + 2 2 i}{2 \cdot 6} : - \frac{2}{3} + i \frac{2}{6}

u = \frac{- 8 - 2 2 i}{2 \cdot 6} : - \frac{2}{3} - i \frac{2}{6}

The solutions to the quadratic equation are:

u = - \frac{2}{3} + i \frac{2}{6}, u = - \frac{2}{3} - i \frac{2}{6}

\frac{1}{25} - x^{2} = 0, (\frac{2}{5}, - \frac{1}{5}, \frac{1}{5}, - \frac{2}{5})

- 5 x^{2} + 2 = x

4 y^{2} + 8 y + 4 = 0

x, z = - x^{2} + y^{2} + 1

3 x^{2} + 1 = 13

What are the solutions to the equation 6u^2+8u=-3 ?
The solutions to the equation 6u^2+8u=-3 are u=-2/3+i(sqrt(2))/6 ,u=-2/3-i(sqrt(2))/6
Find the zeros of 6u^2+8u=-3
The zeros of 6u^2+8u=-3 are u=-2/3+i(sqrt(2))/6 ,u=-2/3-i(sqrt(2))/6