(5u+1)(8-u)=0

Solution for (5u+1)(8-u)=0 equation:

(5u+1)(8-u)=0

We add all the numbers together, and all the variables

(5u+1)(-1u+8)=0

We multiply parentheses ..

(-5u^2+40u-1u+8)=0

We get rid of parentheses

-5u^2+40u-1u+8=0

We add all the numbers together, and all the variables

-5u^2+39u+8=0

a = -5; b = 39; c = +8;
Δ = b2-4ac
Δ = 392-4·(-5)·8
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-41}{2*-5}=\frac{-80}{-10}
=+8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+41}{2*-5}=\frac{2}{-10}
=-1/5
$