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

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

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

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

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

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

Solution steps

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

Switch sides

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

∣ 2 s + 1 ∣ + 3 \leq 2

3 x (5 x + 2) = 0

2 x^{5} + 2 x^{2} 2 y^{3} - 8 x^{3} - 8 y^{3}

5^{4} \cdot 5^{- 2}

lo g_{\frac{1}{5}} (2)

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