What are the solutions to the equation y^2+2y+5y=5y^2-4y ?

The solutions to the equation y^2+2y+5y=5y^2-4y are y=0,y= 11/4

Find the zeros of y^2+2y+5y=5y^2-4y

The zeros of y^2+2y+5y=5y^2-4y are y=0,y= 11/4

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

y = 0, y = \frac{11}{4}

Decimal Notation

y = 0, y = 2.75

Solution steps

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

Refine

y^{2} + 7 y = 5 y^{2} - 4 y

Move

4 y

to the left side

y^{2} + 11 y = 5 y^{2}

Move

5 y^{2}

to the left side

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

Solve with the quadratic formula

y_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

1 1^{2} - 4 (- 4) \cdot 0 = 11

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

Separate the solutions

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

y = \frac{- 11 + 11}{2 ( - 4 )} : 0

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

The solutions to the quadratic equation are:

y = 0, y = \frac{11}{4}

x, 6 x^{2} - 6 x y + 6 x = 0

4 x^{2} - 36 x + 85 = 0

x^{2} - 7 = 9 x

x^{2} + 6 x - 9 = 6 x

3 y^{2} + 2 y - 1 = 0

What are the solutions to the equation y^2+2y+5y=5y^2-4y ?
The solutions to the equation y^2+2y+5y=5y^2-4y are y=0,y= 11/4
Find the zeros of y^2+2y+5y=5y^2-4y
The zeros of y^2+2y+5y=5y^2-4y are y=0,y= 11/4