What are the solutions to the equation 2x^2+8x+16=0 ?

The solutions to the equation 2x^2+8x+16=0 are x=-2+2i,x=-2-2i

Find the zeros of 2x^2+8x+16=0

The zeros of 2x^2+8x+16=0 are x=-2+2i,x=-2-2i

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2 x^{2} + 8 x + 16 = 0

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

Solution steps

2 x^{2} + 8 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 16}{2 \cdot 2}

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} + 2 x - 1680 = 0

3 x^{2} - 6 x - 15 = 12 x + 12

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

(x - 8)^{2} = - 81

12 x^{2} - 88 x + 112 = 0

What are the solutions to the equation 2x^2+8x+16=0 ?
The solutions to the equation 2x^2+8x+16=0 are x=-2+2i,x=-2-2i
Find the zeros of 2x^2+8x+16=0
The zeros of 2x^2+8x+16=0 are x=-2+2i,x=-2-2i