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

Solution for (8-y)(3y-4)=0 equation:

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

We add all the numbers together, and all the variables

(-1y+8)(3y-4)=0

We multiply parentheses ..

(-3y^2+4y+24y-32)=0

We get rid of parentheses

-3y^2+4y+24y-32=0

We add all the numbers together, and all the variables

-3y^2+28y-32=0

a = -3; b = 28; c = -32;
Δ = b2-4ac
Δ = 282-4·(-3)·(-32)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-20}{2*-3}=\frac{-48}{-6}
=+8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+20}{2*-3}=\frac{-8}{-6}
=1+1/3
$