(3y-8)(5-y)=0

Solution for (3y-8)(5-y)=0 equation:

(3y-8)(5-y)=0

We add all the numbers together, and all the variables

(3y-8)(-1y+5)=0

We multiply parentheses ..

(-3y^2+15y+8y-40)=0

We get rid of parentheses

-3y^2+15y+8y-40=0

We add all the numbers together, and all the variables

-3y^2+23y-40=0

a = -3; b = 23; c = -40;
Δ = b2-4ac
Δ = 232-4·(-3)·(-40)
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{-(23)-7}{2*-3}=\frac{-30}{-6}
=+5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*-3}=\frac{-16}{-6}
=2+2/3
$