What are the solutions to the equation x^2+2x+19=0 ?

The solutions to the equation x^2+2x+19=0 are x=-1+3sqrt(2)i,x=-1-3sqrt(2)i

Find the zeros of x^2+2x+19=0

The zeros of x^2+2x+19=0 are x=-1+3sqrt(2)i,x=-1-3sqrt(2)i

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x^{2} + 2 x + 19 = 0

x = - 1 + 32 i, x = - 1 - 32 i

Solution steps

x^{2} + 2 x + 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 19}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 19 : 62 i

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

Separate the solutions

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

x = \frac{- 2 + 6 2 i}{2 \cdot 1} : - 1 + 32 i

x = \frac{- 2 - 6 2 i}{2 \cdot 1} : - 1 - 32 i

The solutions to the quadratic equation are:

x = - 1 + 32 i, x = - 1 - 32 i

0 = 12 x^{2} - 16

x^{2} = \frac{x}{3} + 8

18 x^{2} - 50 = 0

10 x^{2} - 26 x = - 12

x^{2} = 6 x + 91

What are the solutions to the equation x^2+2x+19=0 ?
The solutions to the equation x^2+2x+19=0 are x=-1+3sqrt(2)i,x=-1-3sqrt(2)i
Find the zeros of x^2+2x+19=0
The zeros of x^2+2x+19=0 are x=-1+3sqrt(2)i,x=-1-3sqrt(2)i