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

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

Find the zeros of (x^2+2x+5)=0

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

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

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

Solution steps

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

Refine

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

- 5 x^{2} - 30 x - 405 = 0

x^{2} - 3 x = 6.75

x^{2} - 12 x = - 8

x, x^{2} + 4 y^{2} + 16 = 0

x^{2} + 7 x - 14 = 7 x + 12

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