What are the solutions to the equation 2x^2+6x=-5 ?

The solutions to the equation 2x^2+6x=-5 are x=-3/2+i 1/2 ,x=-3/2-i 1/2

Find the zeros of 2x^2+6x=-5

The zeros of 2x^2+6x=-5 are x=-3/2+i 1/2 ,x=-3/2-i 1/2

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2 x^{2} + 6 x = - 5

x = - \frac{3}{2} + i \frac{1}{2}, x = - \frac{3}{2} - i \frac{1}{2}

Solution steps

2 x^{2} + 6 x = - 5

Move

5

to the left side

2 x^{2} + 6 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

Simplify

6^{2} - 4 \cdot 2 \cdot 5 : 2 i

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

Separate the solutions

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

x = \frac{- 6 + 2 i}{2 \cdot 2} : - \frac{3}{2} + i \frac{1}{2}

x = \frac{- 6 - 2 i}{2 \cdot 2} : - \frac{3}{2} - i \frac{1}{2}

The solutions to the quadratic equation are:

x = - \frac{3}{2} + i \frac{1}{2}, x = - \frac{3}{2} - i \frac{1}{2}

a^{2} - a - 30

\frac{3}{5} (x - \frac{3}{2}) = \frac{27}{10}

(\frac{- 1 + 4 x ^{\frac{2}{3}}}{4 x ^{\frac{1}{3}}})^{2}

176 = 8 (1 x^{3} x)

- 47 = - 7 + 2 x

What are the solutions to the equation 2x^2+6x=-5 ?
The solutions to the equation 2x^2+6x=-5 are x=-3/2+i 1/2 ,x=-3/2-i 1/2
Find the zeros of 2x^2+6x=-5
The zeros of 2x^2+6x=-5 are x=-3/2+i 1/2 ,x=-3/2-i 1/2