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

The solutions to the equation y(4y+1)=8 are y=(-1+sqrt(129))/8 ,y=(-1-sqrt(129))/8

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

The zeros of y(4y+1)=8 are y=(-1+sqrt(129))/8 ,y=(-1-sqrt(129))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y (4 y + 1) = 8

y = \frac{- 1 + 129}{8}, y = \frac{- 1 - 129}{8}

Decimal Notation

y = 1.29472 \dots, y = - 1.54472 \dots

Solution steps

y (4 y + 1) = 8

Expand

y (4 y + 1) : 4 y^{2} + y

4 y^{2} + y = 8

Move

8

to the left side

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 4 (- 8) = 129

y_{1, 2} = \frac{- 1 \pm 129}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 1 + 129}{2 \cdot 4}, y_{2} = \frac{- 1 - 129}{2 \cdot 4}

y = \frac{- 1 + 129}{2 \cdot 4} : \frac{- 1 + 129}{8}

y = \frac{- 1 - 129}{2 \cdot 4} : \frac{- 1 - 129}{8}

The solutions to the quadratic equation are:

y = \frac{- 1 + 129}{8}, y = \frac{- 1 - 129}{8}

5 x^{2} - 100 = 25

x^{2} - 4 = 8

28 x - 49 x^{2} = 4

3 x^{2} + 12 x - 9

x, 9 x^{2} - m^{2} = 0

What are the solutions to the equation y(4y+1)=8 ?
The solutions to the equation y(4y+1)=8 are y=(-1+sqrt(129))/8 ,y=(-1-sqrt(129))/8
Find the zeros of y(4y+1)=8
The zeros of y(4y+1)=8 are y=(-1+sqrt(129))/8 ,y=(-1-sqrt(129))/8