2y(y-4)=y(4-y)

Solution for 2y(y-4)=y(4-y) equation:

2y(y-4)=y(4-y)

We move all terms to the left:

2y(y-4)-(y(4-y))=0

We add all the numbers together, and all the variables

2y(y-4)-(y(-1y+4))=0

We multiply parentheses

2y^2-8y-(y(-1y+4))=0


We calculate terms in parentheses: -(y(-1y+4)), so:

y(-1y+4)

We multiply parentheses

-1y^2+4y

Back to the equation:

-(-1y^2+4y)


We get rid of parentheses

2y^2+1y^2-4y-8y=0

We add all the numbers together, and all the variables

3y^2-12y=0

a = 3; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·3·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*3}=\frac{0}{6}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*3}=\frac{24}{6}
=4
$