y.5(3y-9)=12y+18-9y+9

Solution for y.5(3y-9)=12y+18-9y+9 equation:

y.5(3y-9)=12y+18-9y+9

We move all terms to the left:

y.5(3y-9)-(12y+18-9y+9)=0

We add all the numbers together, and all the variables

y.5(3y-9)-(3y+27)=0

We multiply parentheses

3y^2-9y-(3y+27)=0

We get rid of parentheses

3y^2-9y-3y-27=0

We add all the numbers together, and all the variables

3y^2-12y-27=0

a = 3; b = -12; c = -27;
Δ = b2-4ac
Δ = -122-4·3·(-27)
Δ = 468
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{13}}{2*3}=\frac{12-6\sqrt{13}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{13}}{2*3}=\frac{12+6\sqrt{13}}{6}
$