What are the solutions to the equation 2+4u^2=-3u ?

The solutions to the equation 2+4u^2=-3u are u=-3/8+i(sqrt(23))/8 ,u=-3/8-i(sqrt(23))/8

Find the zeros of 2+4u^2=-3u

The zeros of 2+4u^2=-3u are u=-3/8+i(sqrt(23))/8 ,u=-3/8-i(sqrt(23))/8

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2 + 4 u^{2} = - 3 u

u = - \frac{3}{8} + i \frac{23}{8}, u = - \frac{3}{8} - i \frac{23}{8}

Solution steps

2 + 4 u^{2} = - 3 u

Move

3 u

to the left side

2 + 4 u^{2} + 3 u = 0

Write in the standard form

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

4 u^{2} + 3 u + 2 = 0

Solve with the quadratic formula

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

Simplify

3^{2} - 4 \cdot 4 \cdot 2 : 23 i

u_{1, 2} = \frac{- 3 \pm 23 i}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 3 + 23 i}{2 \cdot 4}, u_{2} = \frac{- 3 - 23 i}{2 \cdot 4}

u = \frac{- 3 + 23 i}{2 \cdot 4} : - \frac{3}{8} + i \frac{23}{8}

u = \frac{- 3 - 23 i}{2 \cdot 4} : - \frac{3}{8} - i \frac{23}{8}

The solutions to the quadratic equation are:

u = - \frac{3}{8} + i \frac{23}{8}, u = - \frac{3}{8} - i \frac{23}{8}

x^{2} - 25 x = 0

(4 x + 16)^{2} + x^{2} = (4 x + 17)^{2}

\frac{x}{10} = \frac{2}{8}

2^{- 4} = \frac{1}{16}

(4 k)^{2}

What are the solutions to the equation 2+4u^2=-3u ?
The solutions to the equation 2+4u^2=-3u are u=-3/8+i(sqrt(23))/8 ,u=-3/8-i(sqrt(23))/8
Find the zeros of 2+4u^2=-3u
The zeros of 2+4u^2=-3u are u=-3/8+i(sqrt(23))/8 ,u=-3/8-i(sqrt(23))/8