What are the solutions to the equation 0=y^2+8y+16 ?

The solutions to the equation 0=y^2+8y+16 are y=-4

Find the zeros of 0=y^2+8y+16

The zeros of 0=y^2+8y+16 are y=-4

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0 = y^{2} + 8 y + 16

y = - 4

Solution steps

0 = y^{2} + 8 y + 16

Switch sides

y^{2} + 8 y + 16 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 16 = 0

y_{1, 2} = \frac{- 8 \pm 0}{2 \cdot 1}

y = \frac{- 8}{2 \cdot 1}

\frac{- 8}{2 \cdot 1} = - 4

y = - 4

The solution to the quadratic equation is:

y = - 4

2 - \frac{5}{6} v \geq v + \frac{3}{8}

735

x^{9} - 8

144 x^{2} + 81 = 0

a^{4} - 1

What are the solutions to the equation 0=y^2+8y+16 ?
The solutions to the equation 0=y^2+8y+16 are y=-4
Find the zeros of 0=y^2+8y+16
The zeros of 0=y^2+8y+16 are y=-4