What are the solutions to the equation 2x^2+40=-8x ?

The solutions to the equation 2x^2+40=-8x are x=-2+4i,x=-2-4i

Find the zeros of 2x^2+40=-8x

The zeros of 2x^2+40=-8x are x=-2+4i,x=-2-4i

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2 x^{2} + 40 = - 8 x

x = - 2 + 4 i, x = - 2 - 4 i

Solution steps

2 x^{2} + 40 = - 8 x

Move

8 x

to the left side

2 x^{2} + 40 + 8 x = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} + 8 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 40}{2 \cdot 2}

Simplify

8^{2} - 4 \cdot 2 \cdot 40 : 16 i

x_{1, 2} = \frac{- 8 \pm 16 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 16 i}{2 \cdot 2}, x_{2} = \frac{- 8 - 16 i}{2 \cdot 2}

x = \frac{- 8 + 16 i}{2 \cdot 2} : - 2 + 4 i

x = \frac{- 8 - 16 i}{2 \cdot 2} : - 2 - 4 i

The solutions to the quadratic equation are:

x = - 2 + 4 i, x = - 2 - 4 i

x^{2} - 8 x = 65

x, x^{2} + 6 x - 4 y^{2} + 8 y + 5 = 0

3 x^{2} - 6 x + 13 = 0

2 x^{2} = 144

3 x^{2} - 24 x + 2 = 5 x^{2} + 1

What are the solutions to the equation 2x^2+40=-8x ?
The solutions to the equation 2x^2+40=-8x are x=-2+4i,x=-2-4i
Find the zeros of 2x^2+40=-8x
The zeros of 2x^2+40=-8x are x=-2+4i,x=-2-4i