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

The solutions to the equation 2w^2+4w=-3 are w=-1+i(sqrt(2))/2 ,w=-1-i(sqrt(2))/2

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

The zeros of 2w^2+4w=-3 are w=-1+i(sqrt(2))/2 ,w=-1-i(sqrt(2))/2

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

w = - 1 + i \frac{2}{2}, w = - 1 - i \frac{2}{2}

Solution steps

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

Move

3

to the left side

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

Solve with the quadratic formula

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

Simplify

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

w_{1, 2} = \frac{- 4 \pm 2 2 i}{2 \cdot 2}

Separate the solutions

w_{1} = \frac{- 4 + 2 2 i}{2 \cdot 2}, w_{2} = \frac{- 4 - 2 2 i}{2 \cdot 2}

w = \frac{- 4 + 2 2 i}{2 \cdot 2} : - 1 + i \frac{2}{2}

w = \frac{- 4 - 2 2 i}{2 \cdot 2} : - 1 - i \frac{2}{2}

The solutions to the quadratic equation are:

w = - 1 + i \frac{2}{2}, w = - 1 - i \frac{2}{2}

p^{2} - 10 p - 52 = 0

y, x^{2} + y^{3} = 4

5 x^{2} = 15 x

4 x^{2} + 57 = - 28 x

3 x^{2} - 136 x + 1300 = 0

What are the solutions to the equation 2w^2+4w=-3 ?
The solutions to the equation 2w^2+4w=-3 are w=-1+i(sqrt(2))/2 ,w=-1-i(sqrt(2))/2
Find the zeros of 2w^2+4w=-3
The zeros of 2w^2+4w=-3 are w=-1+i(sqrt(2))/2 ,w=-1-i(sqrt(2))/2