(2y-3)(y-2)=12y+18

Solution for (2y-3)(y-2)=12y+18 equation:

(2y-3)(y-2)=12y+18

We move all terms to the left:

(2y-3)(y-2)-(12y+18)=0

We get rid of parentheses

(2y-3)(y-2)-12y-18=0

We multiply parentheses ..

(+2y^2-4y-3y+6)-12y-18=0

We get rid of parentheses

2y^2-4y-3y-12y+6-18=0

We add all the numbers together, and all the variables

2y^2-19y-12=0

a = 2; b = -19; c = -12;
Δ = b2-4ac
Δ = -192-4·2·(-12)
Δ = 457
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{-(-19)-\sqrt{457}}{2*2}=\frac{19-\sqrt{457}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{457}}{2*2}=\frac{19+\sqrt{457}}{4}
$