3y(y-5)=2(y+12)

Solution for 3y(y-5)=2(y+12) equation:

3y(y-5)=2(y+12)

We move all terms to the left:

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

We multiply parentheses

3y^2-15y-(2(y+12))=0


We calculate terms in parentheses: -(2(y+12)), so:

2(y+12)

We multiply parentheses

2y+24

Back to the equation:

-(2y+24)


We get rid of parentheses

3y^2-15y-2y-24=0

We add all the numbers together, and all the variables

3y^2-17y-24=0

a = 3; b = -17; c = -24;
Δ = b2-4ac
Δ = -172-4·3·(-24)
Δ = 577
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-\sqrt{577}}{2*3}=\frac{17-\sqrt{577}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{577}}{2*3}=\frac{17+\sqrt{577}}{6}
$