What are the solutions to the equation 2x^2+8x+80=0 ?

The solutions to the equation 2x^2+8x+80=0 are x=-2+6i,x=-2-6i

Find the zeros of 2x^2+8x+80=0

The zeros of 2x^2+8x+80=0 are x=-2+6i,x=-2-6i

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2 x^{2} + 8 x + 80 = 0

x = - 2 + 6 i, x = - 2 - 6 i

Solution steps

2 x^{2} + 8 x + 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 80}{2 \cdot 2}

Simplify

8^{2} - 4 \cdot 2 \cdot 80 : 24 i

x_{1, 2} = \frac{- 8 \pm 24 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 24 i}{2 \cdot 2}, x_{2} = \frac{- 8 - 24 i}{2 \cdot 2}

x = \frac{- 8 + 24 i}{2 \cdot 2} : - 2 + 6 i

x = \frac{- 8 - 24 i}{2 \cdot 2} : - 2 - 6 i

The solutions to the quadratic equation are:

x = - 2 + 6 i, x = - 2 - 6 i

x, r (x)^{2} = x^{2} + y^{2}

x^{2} + 14 x + 45 = 0

x^{2} - 2 x = 124

3 k^{2} - 14 k + 21 = 0

y, x^{2} + y^{2} + 10 x - 8 y + 41 = 0

What are the solutions to the equation 2x^2+8x+80=0 ?
The solutions to the equation 2x^2+8x+80=0 are x=-2+6i,x=-2-6i
Find the zeros of 2x^2+8x+80=0
The zeros of 2x^2+8x+80=0 are x=-2+6i,x=-2-6i