What are the solutions to the equation (20+2y)(y-5)=20y ?

The solutions to the equation (20+2y)(y-5)=20y are y=10,y=-5

Find the zeros of (20+2y)(y-5)=20y

The zeros of (20+2y)(y-5)=20y are y=10,y=-5

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(20 + 2 y) (y - 5) = 20 y

y = 10, y = - 5

Solution steps

(20 + 2 y) (y - 5) = 20 y

Expand

(20 + 2 y) (y - 5) : 2 y^{2} + 10 y - 100

2 y^{2} + 10 y - 100 = 20 y

Move

20 y

to the left side

2 y^{2} - 10 y - 100 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 2 ( - 100 )}{2 \cdot 2}

(- 10)^{2} - 4 \cdot 2 (- 100) = 30

y_{1, 2} = \frac{- ( - 10 ) \pm 30}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 10 ) + 30}{2 \cdot 2}, y_{2} = \frac{- ( - 10 ) - 30}{2 \cdot 2}

y = \frac{- ( - 10 ) + 30}{2 \cdot 2} : 10

y = \frac{- ( - 10 ) - 30}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

y = 10, y = - 5

x, x^{2} + 4 x + 7 = 0

x^{2} - 2 - 3 = 0

x^{2} + 8 x - 64 = 0

3 (x + 1)^{2} - 48 = 0

X, Y = X^{2} - 4 X + 3

What are the solutions to the equation (20+2y)(y-5)=20y ?
The solutions to the equation (20+2y)(y-5)=20y are y=10,y=-5
Find the zeros of (20+2y)(y-5)=20y
The zeros of (20+2y)(y-5)=20y are y=10,y=-5