What are the solutions to the equation 2x^2+6x=-18 ?

The solutions to the equation 2x^2+6x=-18 are x=-3/2+i(3sqrt(3))/2 ,x=-3/2-i(3sqrt(3))/2

Find the zeros of 2x^2+6x=-18

The zeros of 2x^2+6x=-18 are x=-3/2+i(3sqrt(3))/2 ,x=-3/2-i(3sqrt(3))/2

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2 x^{2} + 6 x = - 18

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

Solution steps

2 x^{2} + 6 x = - 18

Move

18

to the left side

2 x^{2} + 6 x + 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 \cdot 18}{2 \cdot 2}

Simplify

6^{2} - 4 \cdot 2 \cdot 18 : 63 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{- 8}{5}

- 2 > - 2

x < 3 and x \geq 5

5 - \frac{x}{12} = 10

x^{4} - 10 x^{3} + 24 x^{2} + 32 x - 128

What are the solutions to the equation 2x^2+6x=-18 ?
The solutions to the equation 2x^2+6x=-18 are x=-3/2+i(3sqrt(3))/2 ,x=-3/2-i(3sqrt(3))/2
Find the zeros of 2x^2+6x=-18
The zeros of 2x^2+6x=-18 are x=-3/2+i(3sqrt(3))/2 ,x=-3/2-i(3sqrt(3))/2