What are the solutions to the equation 4y^2+8y+4=0 ?

The solutions to the equation 4y^2+8y+4=0 are y=-1

Find the zeros of 4y^2+8y+4=0

The zeros of 4y^2+8y+4=0 are y=-1

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4 y^{2} + 8 y + 4 = 0

y = - 1

Solution steps

4 y^{2} + 8 y + 4 = 0

Solve with the quadratic formula

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

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

y_{1, 2} = \frac{- 8 \pm 0}{2 \cdot 4}

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

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

y = - 1

The solution to the quadratic equation is:

y = - 1

x, z = - x^{2} + y^{2} + 1

3 x^{2} + 1 = 13

k^{2} - 18 k - 4 = 41

18 m^{2} - 27 m - 3 = 0

0.5 x^{2} - 1.61 x + 1.4 = 0

What are the solutions to the equation 4y^2+8y+4=0 ?
The solutions to the equation 4y^2+8y+4=0 are y=-1
Find the zeros of 4y^2+8y+4=0
The zeros of 4y^2+8y+4=0 are y=-1