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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

3y(y-3)-(30-2y)+5=0

We multiply parentheses

3y^2-9y-(30-2y)+5=0

We get rid of parentheses

3y^2-9y+2y-30+5=0

We add all the numbers together, and all the variables

3y^2-7y-25=0

a = 3; b = -7; c = -25;
Δ = b2-4ac
Δ = -72-4·3·(-25)
Δ = 349
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{-(-7)-\sqrt{349}}{2*3}=\frac{7-\sqrt{349}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{349}}{2*3}=\frac{7+\sqrt{349}}{6}
$