(5u+3)(2-u)=0

Solution for (5u+3)(2-u)=0 equation:

(5u+3)(2-u)=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-5u^2+10u-3u+6)=0

We get rid of parentheses

-5u^2+10u-3u+6=0

We add all the numbers together, and all the variables

-5u^2+7u+6=0

a = -5; b = 7; c = +6;
Δ = b2-4ac
Δ = 72-4·(-5)·6
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*-5}=\frac{-20}{-10}
=+2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*-5}=\frac{6}{-10}
=-3/5
$