What are the solutions to the equation x^2+2*x+5=0 ?

The solutions to the equation x^2+2*x+5=0 are x=-1+2i,x=-1-2i

Find the zeros of x^2+2*x+5=0

The zeros of x^2+2*x+5=0 are x=-1+2i,x=-1-2i

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x^{2} + 2 \cdot x + 5 = 0

x = - 1 + 2 i, x = - 1 - 2 i

Solution steps

x^{2} + 2 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 5 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1 + 2 i, x = - 1 - 2 i

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

16 x^{2} - 256 = 0

0.003 x^{2} - 0.04 x + 2 - a = 0

4.9 t^{2} - 20 t + 15 = 0

21 - 8 k - 2 k^{2} = 2 k^{2}

What are the solutions to the equation x^2+2*x+5=0 ?
The solutions to the equation x^2+2*x+5=0 are x=-1+2i,x=-1-2i
Find the zeros of x^2+2*x+5=0
The zeros of x^2+2*x+5=0 are x=-1+2i,x=-1-2i