(5u+1)(4-u)=0

Solution for (5u+1)(4-u)=0 equation:

(5u+1)(4-u)=0

We add all the numbers together, and all the variables

(5u+1)(-1u+4)=0

We multiply parentheses ..

(-5u^2+20u-1u+4)=0

We get rid of parentheses

-5u^2+20u-1u+4=0

We add all the numbers together, and all the variables

-5u^2+19u+4=0

a = -5; b = 19; c = +4;
Δ = b2-4ac
Δ = 192-4·(-5)·4
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*-5}=\frac{-40}{-10}
=+4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*-5}=\frac{2}{-10}
=-1/5
$