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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3u^2+9u+5u-15)=0

We get rid of parentheses

-3u^2+9u+5u-15=0

We add all the numbers together, and all the variables

-3u^2+14u-15=0

a = -3; b = 14; c = -15;
Δ = b2-4ac
Δ = 142-4·(-3)·(-15)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*-3}=\frac{-18}{-6}
=+3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*-3}=\frac{-10}{-6}
=1+2/3
$