What are the solutions to the equation r^2+2r+17=0 ?

The solutions to the equation r^2+2r+17=0 are r=-1+4i,r=-1-4i

Find the zeros of r^2+2r+17=0

The zeros of r^2+2r+17=0 are r=-1+4i,r=-1-4i

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r^{2} + 2 r + 17 = 0

r = - 1 + 4 i, r = - 1 - 4 i

Solution steps

r^{2} + 2 r + 17 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 17}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 17 : 8 i

r_{1, 2} = \frac{- 2 \pm 8 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 2 + 8 i}{2 \cdot 1}, r_{2} = \frac{- 2 - 8 i}{2 \cdot 1}

r = \frac{- 2 + 8 i}{2 \cdot 1} : - 1 + 4 i

r = \frac{- 2 - 8 i}{2 \cdot 1} : - 1 - 4 i

The solutions to the quadratic equation are:

r = - 1 + 4 i, r = - 1 - 4 i

x^{2} + 3 x - 5 = 0

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16.8 = \frac{x}{3.09}

n^{2} + 14 n + 13

What are the solutions to the equation r^2+2r+17=0 ?
The solutions to the equation r^2+2r+17=0 are r=-1+4i,r=-1-4i
Find the zeros of r^2+2r+17=0
The zeros of r^2+2r+17=0 are r=-1+4i,r=-1-4i