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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4y^2+2y-12=0

a = 4; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·4·(-12)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*4}=\frac{-16}{8}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*4}=\frac{12}{8}
=1+1/2
$