What are the solutions to the equation y^2+2y+10=0 ?

The solutions to the equation y^2+2y+10=0 are y=-1+3i,y=-1-3i

Find the zeros of y^2+2y+10=0

The zeros of y^2+2y+10=0 are y=-1+3i,y=-1-3i

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

y = - 1 + 3 i, y = - 1 - 3 i

Solution steps

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

Solve with the quadratic formula

y_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 1 + 3 i, y = - 1 - 3 i

0 - \frac{1}{2}

(x + y)^{- 1}

(x^{2} + 4 x + 8)^{2} + 3 x (x^{2} + 4 x + 8) + 2 x^{2}

- 3 (\frac{8}{9})^{0}

(x^{4})^{- 5}

What are the solutions to the equation y^2+2y+10=0 ?
The solutions to the equation y^2+2y+10=0 are y=-1+3i,y=-1-3i
Find the zeros of y^2+2y+10=0
The zeros of y^2+2y+10=0 are y=-1+3i,y=-1-3i