3y(2y+3)=4y+6

Solution for 3y(2y+3)=4y+6 equation:

3y(2y+3)=4y+6

We move all terms to the left:

3y(2y+3)-(4y+6)=0

We multiply parentheses

6y^2+9y-(4y+6)=0

We get rid of parentheses

6y^2+9y-4y-6=0

We add all the numbers together, and all the variables

6y^2+5y-6=0

a = 6; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·6·(-6)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*6}=\frac{-18}{12}
=-1+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*6}=\frac{8}{12}
=2/3
$