What are the solutions to the equation y(4y-5)=6y+1 ?

The solutions to the equation y(4y-5)=6y+1 are y=(11+sqrt(137))/8 ,y=(11-sqrt(137))/8

Find the zeros of y(4y-5)=6y+1

The zeros of y(4y-5)=6y+1 are y=(11+sqrt(137))/8 ,y=(11-sqrt(137))/8

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y (4 y - 5) = 6 y + 1

y = \frac{11 + 137}{8}, y = \frac{11 - 137}{8}

Decimal Notation

y = 2.83808 \dots, y = - 0.08808 \dots

Solution steps

y (4 y - 5) = 6 y + 1

Expand

y (4 y - 5) : 4 y^{2} - 5 y

4 y^{2} - 5 y = 6 y + 1

Move

1

to the left side

4 y^{2} - 5 y - 1 = 6 y

Move

6 y

to the left side

4 y^{2} - 11 y - 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 ( - 1 )}{2 \cdot 4}

(- 11)^{2} - 4 \cdot 4 (- 1) = 137

y_{1, 2} = \frac{- ( - 11 ) \pm 137}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 11 ) + 137}{2 \cdot 4}, y_{2} = \frac{- ( - 11 ) - 137}{2 \cdot 4}

y = \frac{- ( - 11 ) + 137}{2 \cdot 4} : \frac{11 + 137}{8}

y = \frac{- ( - 11 ) - 137}{2 \cdot 4} : \frac{11 - 137}{8}

The solutions to the quadratic equation are:

y = \frac{11 + 137}{8}, y = \frac{11 - 137}{8}

2 p^{2} - 16 p - 13 = 0

x, y^{2} - 2 y = x^{2} - x + c

2 r^{2} - 6 r - 1 = - 3 r

5 x^{2} + 13 x + 30 = 0

7 x^{2} + 56 x = 0

What are the solutions to the equation y(4y-5)=6y+1 ?
The solutions to the equation y(4y-5)=6y+1 are y=(11+sqrt(137))/8 ,y=(11-sqrt(137))/8
Find the zeros of y(4y-5)=6y+1
The zeros of y(4y-5)=6y+1 are y=(11+sqrt(137))/8 ,y=(11-sqrt(137))/8