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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2y^2-7y=0

a = 2; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·2·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*2}=\frac{0}{4}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*2}=\frac{14}{4}
=3+1/2
$