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

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

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

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

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

y = - 1 + 4 i, y = - 1 - 4 i

Solution steps

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

Solve with the quadratic formula

y_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 1 + 4 i, y = - 1 - 4 i

2 x^{2} + 17 x - 9 = 0

- 3 x x_{+} 5 (x - 2)

- 2 x = 10

\frac{4}{3} (6)

\frac{a - 7}{a ^{2} - 3 a - 28} \cdot \frac{8 a - 40}{a ^{2} + 3 a - 40}

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