What are the solutions to the equation 2y(8y+3)=1 ?

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

Find the zeros of 2y(8y+3)=1

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

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2 y (8 y + 3) = 1

y = \frac{1}{8}, y = - \frac{1}{2}

Decimal Notation

y = 0.125, y = - 0.5

Solution steps

2 y (8 y + 3) = 1

Expand

2 y (8 y + 3) : 16 y^{2} + 6 y

16 y^{2} + 6 y = 1

Move

1

to the left side

16 y^{2} + 6 y - 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 16 ( - 1 )}{2 \cdot 16}

6^{2} - 4 \cdot 16 (- 1) = 10

y_{1, 2} = \frac{- 6 \pm 10}{2 \cdot 16}

Separate the solutions

y_{1} = \frac{- 6 + 10}{2 \cdot 16}, y_{2} = \frac{- 6 - 10}{2 \cdot 16}

y = \frac{- 6 + 10}{2 \cdot 16} : \frac{1}{8}

y = \frac{- 6 - 10}{2 \cdot 16} : - \frac{1}{2}

The solutions to the quadratic equation are:

y = \frac{1}{8}, y = - \frac{1}{2}

0 = 5 - \frac{1}{2} x^{2}

(2 x - 7) (4 x + 3) = 112

u^{2} - 15 u - 34 = 0

a, a^{2} b - c = d

4 k^{2} - 64 = 0

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