(3-y)(3y-4)=0

Solution for (3-y)(3y-4)=0 equation:

(3-y)(3y-4)=0

We add all the numbers together, and all the variables

(-1y+3)(3y-4)=0

We multiply parentheses ..

(-3y^2+4y+9y-12)=0

We get rid of parentheses

-3y^2+4y+9y-12=0

We add all the numbers together, and all the variables

-3y^2+13y-12=0

a = -3; b = 13; c = -12;
Δ = b2-4ac
Δ = 132-4·(-3)·(-12)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*-3}=\frac{-18}{-6}
=+3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*-3}=\frac{-8}{-6}
=1+1/3
$