(4-u)(5u+3)=0

Solution for (4-u)(5u+3)=0 equation:

(4-u)(5u+3)=0

We add all the numbers together, and all the variables

(-1u+4)(5u+3)=0

We multiply parentheses ..

(-5u^2-3u+20u+12)=0

We get rid of parentheses

-5u^2-3u+20u+12=0

We add all the numbers together, and all the variables

-5u^2+17u+12=0

a = -5; b = 17; c = +12;
Δ = b2-4ac
Δ = 172-4·(-5)·12
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-23}{2*-5}=\frac{-40}{-10}
=+4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+23}{2*-5}=\frac{6}{-10}
=-3/5
$