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

Solution for (3u+1)(5-u)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3u^2+15u-1u+5)=0

We get rid of parentheses

-3u^2+15u-1u+5=0

We add all the numbers together, and all the variables

-3u^2+14u+5=0

a = -3; b = 14; c = +5;
Δ = b2-4ac
Δ = 142-4·(-3)·5
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-16}{2*-3}=\frac{-30}{-6}
=+5
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+16}{2*-3}=\frac{2}{-6}
=-1/3
$