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

The solutions to the equation x^2+2x+5=-5 are x=-1+3i,x=-1-3i

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

The zeros of x^2+2x+5=-5 are x=-1+3i,x=-1-3i

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

x = - 1 + 3 i, x = - 1 - 3 i

Solution steps

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

Move

5

to the left side

x^{2} + 2 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 10 : 6 i

x_{1, 2} = \frac{- 2 \pm 6 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 6 i}{2 \cdot 1}, x_{2} = \frac{- 2 - 6 i}{2 \cdot 1}

x = \frac{- 2 + 6 i}{2 \cdot 1} : - 1 + 3 i

x = \frac{- 2 - 6 i}{2 \cdot 1} : - 1 - 3 i

The solutions to the quadratic equation are:

x = - 1 + 3 i, x = - 1 - 3 i

x, f (x + 1) = 2 x^{2} + x - 1

63 (p^{2} - 1) = 32 p

- 5 x^{2} - 67 x - 100 = - 7 x

x^{2} + 3 x + 6

a x^{2} + (- 2 a^{2} + 2) x - 4 a = 0

What are the solutions to the equation x^2+2x+5=-5 ?
The solutions to the equation x^2+2x+5=-5 are x=-1+3i,x=-1-3i
Find the zeros of x^2+2x+5=-5
The zeros of x^2+2x+5=-5 are x=-1+3i,x=-1-3i