.(3y-5)+(5y+20)=135

Solution for .(3y-5)+(5y+20)=135 equation:

.(3y-5)+(5y+20)=135

We move all terms to the left:

.(3y-5)+(5y+20)-(135)=0

We multiply parentheses

3y^2+(5y+20)-135+=0

We get rid of parentheses

3y^2+5y+20-135+=0

We add all the numbers together, and all the variables

3y^2+5y=0

a = 3; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·3·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*3}=\frac{-10}{6}
=-1+2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*3}=\frac{0}{6}
=0
$