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 ( - 4 ) ( - 4 )}{2 ( - 4 )}

8^{2} - 4 (- 4) (- 4) = 0

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

y = \frac{- 8}{2 ( - 4 )}

\frac{- 8}{2 ( - 4 )} = 1

y = 1

The solution to the quadratic equation is:

y = 1

\frac{1}{4} (3 x - 8) = 25

2 y^{2} + 6 y = - 2

6 x^{2} + 12 - 90

2 x^{2} + x y - 6 y^{2}

x^{2} = 6 x + 16

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