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

Solution for (8-u)(3u+1)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3u^2-1u+24u+8)=0

We get rid of parentheses

-3u^2-1u+24u+8=0

We add all the numbers together, and all the variables

-3u^2+23u+8=0

a = -3; b = 23; c = +8;
Δ = b2-4ac
Δ = 232-4·(-3)·8
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*-3}=\frac{-48}{-6}
=+8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*-3}=\frac{2}{-6}
=-1/3
$