3y(y+3)/5=2y+6

Solution for 3y(y+3)/5=2y+6 equation:

3y(y+3)/5=2y+6

We move all terms to the left:

3y(y+3)/5-(2y+6)=0

We get rid of parentheses

3y(y+3)/5-2y-6=0

We multiply all the terms by the denominator


3y(y+3)-2y*5-6*5=0

We add all the numbers together, and all the variables

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

We multiply parentheses

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

Wy multiply elements

3y^2+9y-10y-30=0

We add all the numbers together, and all the variables

3y^2-1y-30=0

a = 3; b = -1; c = -30;
Δ = b2-4ac
Δ = -12-4·3·(-30)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-19}{2*3}=\frac{-18}{6}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+19}{2*3}=\frac{20}{6}
=3+1/3
$