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

Solution for (3-y)(3y-2)=0 equation:

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

We add all the numbers together, and all the variables

(-1y+3)(3y-2)=0

We multiply parentheses ..

(-3y^2+2y+9y-6)=0

We get rid of parentheses

-3y^2+2y+9y-6=0

We add all the numbers together, and all the variables

-3y^2+11y-6=0

a = -3; b = 11; c = -6;
Δ = b2-4ac
Δ = 112-4·(-3)·(-6)
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{-(11)-7}{2*-3}=\frac{-18}{-6}
=+3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*-3}=\frac{-4}{-6}
=2/3
$