What are the solutions to the equation 4(-2y+1)8y=-8 ?

The solutions to the equation 4(-2y+1)8y=-8 are y=-(-1+sqrt(3))/4 ,y=(1+sqrt(3))/4

Find the zeros of 4(-2y+1)8y=-8

The zeros of 4(-2y+1)8y=-8 are y=-(-1+sqrt(3))/4 ,y=(1+sqrt(3))/4

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4 (- 2 y + 1) 8 y = - 8

y = - \frac{- 1 + 3}{4}, y = \frac{1 + 3}{4}

Decimal Notation

y = - 0.18301 \dots, y = 0.68301 \dots

Solution steps

4 (- 2 y + 1) \cdot 8 y = - 8

Expand

4 (- 2 y + 1) \cdot 8 y : - 64 y^{2} + 32 y

- 64 y^{2} + 32 y = - 8

Move

8

to the left side

- 64 y^{2} + 32 y + 8 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 64 ) \cdot 8}{2 ( - 64 )}

3 2^{2} - 4 (- 64) \cdot 8 = 323

y_{1, 2} = \frac{- 32 \pm 32 3}{2 ( - 64 )}

Separate the solutions

y_{1} = \frac{- 32 + 32 3}{2 ( - 64 )}, y_{2} = \frac{- 32 - 32 3}{2 ( - 64 )}

y = \frac{- 32 + 32 3}{2 ( - 64 )} : - \frac{- 1 + 3}{4}

y = \frac{- 32 - 32 3}{2 ( - 64 )} : \frac{1 + 3}{4}

The solutions to the quadratic equation are:

y = - \frac{- 1 + 3}{4}, y = \frac{1 + 3}{4}

3 = 5 (x^{2} + 2 x)

F = m v^{2}

0 = - 6 x^{2} + 54

- r^{2} - 18 = 3 r - 2 r^{2}

x^{2} - 2 x + 49 = 11 x + 7

What are the solutions to the equation 4(-2y+1)8y=-8 ?
The solutions to the equation 4(-2y+1)8y=-8 are y=-(-1+sqrt(3))/4 ,y=(1+sqrt(3))/4
Find the zeros of 4(-2y+1)8y=-8
The zeros of 4(-2y+1)8y=-8 are y=-(-1+sqrt(3))/4 ,y=(1+sqrt(3))/4