What are the solutions to the equation 2x^2+2+5x=-4-x ?

The solutions to the equation 2x^2+2+5x=-4-x are x=-3/2+i(sqrt(3))/2 ,x=-3/2-i(sqrt(3))/2

Find the zeros of 2x^2+2+5x=-4-x

The zeros of 2x^2+2+5x=-4-x are x=-3/2+i(sqrt(3))/2 ,x=-3/2-i(sqrt(3))/2

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2 x^{2} + 2 + 5 x = - 4 - x

x = - \frac{3}{2} + i \frac{3}{2}, x = - \frac{3}{2} - i \frac{3}{2}

Solution steps

2 x^{2} + 2 + 5 x = - 4 - x

Move

x

to the left side

2 x^{2} + 2 + 6 x = - 4

Move

4

to the left side

2 x^{2} + 6 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 \cdot 6}{2 \cdot 2}

Simplify

6^{2} - 4 \cdot 2 \cdot 6 : 23 i

x_{1, 2} = \frac{- 6 \pm 2 3 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 6 + 2 3 i}{2 \cdot 2}, x_{2} = \frac{- 6 - 2 3 i}{2 \cdot 2}

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

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

The solutions to the quadratic equation are:

x = - \frac{3}{2} + i \frac{3}{2}, x = - \frac{3}{2} - i \frac{3}{2}

2 b + 4 = - 18 - 9 b

1 - a + 6 a > 6

\frac{4}{6.02 \cdot 1 0 ^{23}}

- 8 \cdot (- 10) - 7 \cdot \frac{1}{- 1}

v + \frac{7}{8} = - \frac{1}{2}

What are the solutions to the equation 2x^2+2+5x=-4-x ?
The solutions to the equation 2x^2+2+5x=-4-x are x=-3/2+i(sqrt(3))/2 ,x=-3/2-i(sqrt(3))/2
Find the zeros of 2x^2+2+5x=-4-x
The zeros of 2x^2+2+5x=-4-x are x=-3/2+i(sqrt(3))/2 ,x=-3/2-i(sqrt(3))/2