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

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

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

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

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

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

Solution steps

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

Move

3

to the left side

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

Solve with the quadratic formula

x_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} - x = 6 x - 6

x x

x + 3 x

2 x + 3 \cdot 0 = 5 x - 0

(- 4 x^{5} y^{3})^{3}

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