(3y-8)(3-y)=0

Solution for (3y-8)(3-y)=0 equation:

(3y-8)(3-y)=0

We add all the numbers together, and all the variables

(3y-8)(-1y+3)=0

We multiply parentheses ..

(-3y^2+9y+8y-24)=0

We get rid of parentheses

-3y^2+9y+8y-24=0

We add all the numbers together, and all the variables

-3y^2+17y-24=0

a = -3; b = 17; c = -24;
Δ = b2-4ac
Δ = 172-4·(-3)·(-24)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-1}{2*-3}=\frac{-18}{-6}
=+3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*-3}=\frac{-16}{-6}
=2+2/3
$