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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3u^2+2u+15u-10)=0

We get rid of parentheses

-3u^2+2u+15u-10=0

We add all the numbers together, and all the variables

-3u^2+17u-10=0

a = -3; b = 17; c = -10;
Δ = b2-4ac
Δ = 172-4·(-3)·(-10)
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{-(17)-13}{2*-3}=\frac{-30}{-6}
=+5
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+13}{2*-3}=\frac{-4}{-6}
=2/3
$