What are the solutions to the equation 2x^2+16x+80=0 ?

The solutions to the equation 2x^2+16x+80=0 are x=-4+2sqrt(6)i,x=-4-2sqrt(6)i

Find the zeros of 2x^2+16x+80=0

The zeros of 2x^2+16x+80=0 are x=-4+2sqrt(6)i,x=-4-2sqrt(6)i

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2 x^{2} + 16 x + 80 = 0

x = - 4 + 26 i, x = - 4 - 26 i

Solution steps

2 x^{2} + 16 x + 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 2 \cdot 80}{2 \cdot 2}

Simplify

1 6^{2} - 4 \cdot 2 \cdot 80 : 86 i

x_{1, 2} = \frac{- 16 \pm 8 6 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 16 + 8 6 i}{2 \cdot 2}, x_{2} = \frac{- 16 - 8 6 i}{2 \cdot 2}

x = \frac{- 16 + 8 6 i}{2 \cdot 2} : - 4 + 26 i

x = \frac{- 16 - 8 6 i}{2 \cdot 2} : - 4 - 26 i

The solutions to the quadratic equation are:

x = - 4 + 26 i, x = - 4 - 26 i

(9^{6} \cdot 7^{- 9})^{- 4}

2 π x + 2 π y + 2 π z

x + 2 < - 4

6 (m - 5) + 3 < 39

\frac{x}{9} \leq 4

What are the solutions to the equation 2x^2+16x+80=0 ?
The solutions to the equation 2x^2+16x+80=0 are x=-4+2sqrt(6)i,x=-4-2sqrt(6)i
Find the zeros of 2x^2+16x+80=0
The zeros of 2x^2+16x+80=0 are x=-4+2sqrt(6)i,x=-4-2sqrt(6)i