What are the solutions to the equation 2x^2+4x+8=0 ?

The solutions to the equation 2x^2+4x+8=0 are x=-1+sqrt(3)i,x=-1-sqrt(3)i

Find the zeros of 2x^2+4x+8=0

The zeros of 2x^2+4x+8=0 are x=-1+sqrt(3)i,x=-1-sqrt(3)i

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

x = - 1 + 3 i, x = - 1 - 3 i

Solution steps

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

Solve with the quadratic formula

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

Simplify

4^{2} - 4 \cdot 2 \cdot 8 : 43 i

x_{1, 2} = \frac{- 4 \pm 4 3 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 4 3 i}{2 \cdot 2}, x_{2} = \frac{- 4 - 4 3 i}{2 \cdot 2}

x = \frac{- 4 + 4 3 i}{2 \cdot 2} : - 1 + 3 i

x = \frac{- 4 - 4 3 i}{2 \cdot 2} : - 1 - 3 i

The solutions to the quadratic equation are:

x = - 1 + 3 i, x = - 1 - 3 i

\frac{9 - i}{- 4 + 5 i}

\frac{( x ^{2} + y ^{2} )}{x ^{2} - y ^{2}} + \frac{y}{x + y} - \frac{x}{x - y}

\frac{n ^{2} + 2 n - 24}{n ^{2} - 11 n + 28}

- 9 x^{2} - 6 x - 1 = 0

\frac{4}{3} \cdot π

What are the solutions to the equation 2x^2+4x+8=0 ?
The solutions to the equation 2x^2+4x+8=0 are x=-1+sqrt(3)i,x=-1-sqrt(3)i
Find the zeros of 2x^2+4x+8=0
The zeros of 2x^2+4x+8=0 are x=-1+sqrt(3)i,x=-1-sqrt(3)i