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

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

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

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

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

s = - 1 + 4 i, s = - 1 - 4 i

Solution steps

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

Solve with the quadratic formula

s_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

s = - 1 + 4 i, s = - 1 - 4 i

x^{2} - 8 x + 17 = 2

0 = x^{2} - 10 x + 30

2 x^{2} - 13 x = 24

- \frac{2}{5} t > - 4 and 0.88 \geq 0.44 t

p^{2} + 16 p + 18 = - 37

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