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

The solutions to the equation 8y^2+8y+2=0 are y=-1/2

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

The zeros of 8y^2+8y+2=0 are y=-1/2

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

y = - \frac{1}{2}

Decimal Notation

y = - 0.5

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

y = - \frac{1}{2}

The solution to the quadratic equation is:

y = - \frac{1}{2}

s^{2} + 16 s + 65 = 0

(x - 4)^{2} - 28 = 8

- 4.9 t^{2} + 11.43 t + 15 = 0

(x - 2)^{2} = - 5

(x - 3)^{2} + 4 = 14

What are the solutions to the equation 8y^2+8y+2=0 ?
The solutions to the equation 8y^2+8y+2=0 are y=-1/2
Find the zeros of 8y^2+8y+2=0
The zeros of 8y^2+8y+2=0 are y=-1/2