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

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

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

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

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

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

Solution steps

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

Switch sides

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)^{2} = a

- x^{2} + 6 x = x

2 x^{2} - 10 = - x

3 (2 b + 3)^{2} = 36

(x - a)^{2} = b^{2}

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