What are the solutions to the equation 2x^2+8x=-32 ?

The solutions to the equation 2x^2+8x=-32 are x=-2+2sqrt(3)i,x=-2-2sqrt(3)i

Find the zeros of 2x^2+8x=-32

The zeros of 2x^2+8x=-32 are x=-2+2sqrt(3)i,x=-2-2sqrt(3)i

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2 x^{2} + 8 x = - 32

x = - 2 + 23 i, x = - 2 - 23 i

Solution steps

2 x^{2} + 8 x = - 32

Move

32

to the left side

2 x^{2} + 8 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 32}{2 \cdot 2}

Simplify

8^{2} - 4 \cdot 2 \cdot 32 : 83 i

x_{1, 2} = \frac{- 8 \pm 8 3 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 8 3 i}{2 \cdot 2}, x_{2} = \frac{- 8 - 8 3 i}{2 \cdot 2}

x = \frac{- 8 + 8 3 i}{2 \cdot 2} : - 2 + 23 i

x = \frac{- 8 - 8 3 i}{2 \cdot 2} : - 2 - 23 i

The solutions to the quadratic equation are:

x = - 2 + 23 i, x = - 2 - 23 i

2 x^{2} + 3 x - 560 = 0

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

x^{2} + 10 x - 1 = 0

2 (x + 1)^{2} + 7 = 1

m^{2} + 4 m + 8 = 0

What are the solutions to the equation 2x^2+8x=-32 ?
The solutions to the equation 2x^2+8x=-32 are x=-2+2sqrt(3)i,x=-2-2sqrt(3)i
Find the zeros of 2x^2+8x=-32
The zeros of 2x^2+8x=-32 are x=-2+2sqrt(3)i,x=-2-2sqrt(3)i