What are the solutions to the equation y^2+18y+81=0 ?

The solutions to the equation y^2+18y+81=0 are y=-9

Find the zeros of y^2+18y+81=0

The zeros of y^2+18y+81=0 are y=-9

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y^{2} + 18 y + 81 = 0

y = - 9

Solution steps

y^{2} + 18 y + 81 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 1 \cdot 81}{2 \cdot 1}

1 8^{2} - 4 \cdot 1 \cdot 81 = 0

y_{1, 2} = \frac{- 18 \pm 0}{2 \cdot 1}

y = \frac{- 18}{2 \cdot 1}

\frac{- 18}{2 \cdot 1} = - 9

y = - 9

The solution to the quadratic equation is:

y = - 9

(125 x^{9} y^{6})^{- \frac{5}{3}}

(2 - 5 i) (2 + 5 i)

m > \frac{1}{2} (11 + 18)

ln (e^{- 0.225}) - 3

∣ 4 x + 5 ∣ \geq x + 11

What are the solutions to the equation y^2+18y+81=0 ?
The solutions to the equation y^2+18y+81=0 are y=-9
Find the zeros of y^2+18y+81=0
The zeros of y^2+18y+81=0 are y=-9