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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3u^2+15u-4u+20)=0

We get rid of parentheses

-3u^2+15u-4u+20=0

We add all the numbers together, and all the variables

-3u^2+11u+20=0

a = -3; b = 11; c = +20;
Δ = b2-4ac
Δ = 112-4·(-3)·20
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*-3}=\frac{-30}{-6}
=+5
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*-3}=\frac{8}{-6}
=-1+1/3
$