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

The solutions to the equation x^2+4x+8=0 are x=-2+2i,x=-2-2i

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

The zeros of x^2+4x+8=0 are x=-2+2i,x=-2-2i

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

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

Solution steps

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

Solve with the quadratic formula

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

Simplify

4^{2} - 4 \cdot 1 \cdot 8 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

1 \leq \frac{4 x - 5}{- 2} < 9

\frac{3 x - 5}{4} = \frac{x + 1}{2}

1 > \frac{q}{2} - 1

\frac{m ^{2}}{100} = \frac{1}{2}

3 \cdot 9

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