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

Solution for (8-y)(3y-2)=0 equation:

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

We add all the numbers together, and all the variables

(-1y+8)(3y-2)=0

We multiply parentheses ..

(-3y^2+2y+24y-16)=0

We get rid of parentheses

-3y^2+2y+24y-16=0

We add all the numbers together, and all the variables

-3y^2+26y-16=0

a = -3; b = 26; c = -16;
Δ = b2-4ac
Δ = 262-4·(-3)·(-16)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-22}{2*-3}=\frac{-48}{-6}
=+8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+22}{2*-3}=\frac{-4}{-6}
=2/3
$