What are the solutions to the equation y^2+4y+29=0 ?

The solutions to the equation y^2+4y+29=0 are y=-2+5i,y=-2-5i

Find the zeros of y^2+4y+29=0

The zeros of y^2+4y+29=0 are y=-2+5i,y=-2-5i

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y^{2} + 4 y + 29 = 0

y = - 2 + 5 i, y = - 2 - 5 i

Solution steps

y^{2} + 4 y + 29 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 29 : 10 i

y_{1, 2} = \frac{- 4 \pm 10 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 4 + 10 i}{2 \cdot 1}, y_{2} = \frac{- 4 - 10 i}{2 \cdot 1}

y = \frac{- 4 + 10 i}{2 \cdot 1} : - 2 + 5 i

y = \frac{- 4 - 10 i}{2 \cdot 1} : - 2 - 5 i

The solutions to the quadratic equation are:

y = - 2 + 5 i, y = - 2 - 5 i

x^{2} + x + 3 > 0

e^{2} = y

2 \cdot 2 + 22

x^{3} + 7 x^{2} > 0

\frac{x ^{\frac{3}{5}}}{x ^{\frac{3}{10}}}

What are the solutions to the equation y^2+4y+29=0 ?
The solutions to the equation y^2+4y+29=0 are y=-2+5i,y=-2-5i
Find the zeros of y^2+4y+29=0
The zeros of y^2+4y+29=0 are y=-2+5i,y=-2-5i