What are the solutions to the equation 2x^2+6x=-9 ?

The solutions to the equation 2x^2+6x=-9 are x=-3/2+i 3/2 ,x=-3/2-i 3/2

Find the zeros of 2x^2+6x=-9

The zeros of 2x^2+6x=-9 are x=-3/2+i 3/2 ,x=-3/2-i 3/2

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2 x^{2} + 6 x = - 9

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

Solution steps

2 x^{2} + 6 x = - 9

Move

9

to the left side

2 x^{2} + 6 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 \cdot 9}{2 \cdot 2}

Simplify

6^{2} - 4 \cdot 2 \cdot 9 : 6 i

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

Separate the solutions

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

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

x = \frac{- 6 - 6 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}

\frac{9}{36}

x^{2} - 9 x - 10 = 0

176 t - 16 t^{2} = 0

\frac{5}{40}

- \frac{15}{3}

What are the solutions to the equation 2x^2+6x=-9 ?
The solutions to the equation 2x^2+6x=-9 are x=-3/2+i 3/2 ,x=-3/2-i 3/2
Find the zeros of 2x^2+6x=-9
The zeros of 2x^2+6x=-9 are x=-3/2+i 3/2 ,x=-3/2-i 3/2