3y(y-4)=5y(y+8)

Solution for 3y(y-4)=5y(y+8) equation:

3y(y-4)=5y(y+8)

We move all terms to the left:

3y(y-4)-(5y(y+8))=0

We multiply parentheses

3y^2-12y-(5y(y+8))=0


We calculate terms in parentheses: -(5y(y+8)), so:

5y(y+8)

We multiply parentheses

5y^2+40y

Back to the equation:

-(5y^2+40y)


We get rid of parentheses

3y^2-5y^2-12y-40y=0

We add all the numbers together, and all the variables

-2y^2-52y=0

a = -2; b = -52; c = 0;
Δ = b2-4ac
Δ = -522-4·(-2)·0
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-52}{2*-2}=\frac{0}{-4}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+52}{2*-2}=\frac{104}{-4}
=-26
$