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

The solutions to the equation 4y^2-2y-2=0 are y=1,y=-1/2

Find the zeros of 4y^2-2y-2=0

The zeros of 4y^2-2y-2=0 are y=1,y=-1/2

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4 y^{2} - 2 y - 2 = 0

y = 1, y = - \frac{1}{2}

Solution steps

4 y^{2} - 2 y - 2 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 4 (- 2) = 6

y_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 4}, y_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 4}

y = \frac{- ( - 2 ) + 6}{2 \cdot 4} : 1

y = \frac{- ( - 2 ) - 6}{2 \cdot 4} : - \frac{1}{2}

The solutions to the quadratic equation are:

y = 1, y = - \frac{1}{2}

(\frac{x}{2})^{2}

7 u + 13 = 3 u + 9

\frac{8 2}{2}

- 3 - 2 i x + i x + (x + 2) 2 i = 0

7 (3 x + 1) + 8 (2 x - 3) = 4 (3 x - 1) - 7 (x - 4)

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