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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4u^2+35u-24=0

a = -4; b = 35; c = -24;
Δ = b2-4ac
Δ = 352-4·(-4)·(-24)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-29}{2*-4}=\frac{-64}{-8}
=+8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+29}{2*-4}=\frac{-6}{-8}
=3/4
$