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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-1u^2+3u+2u-6)=0

We get rid of parentheses

-1u^2+3u+2u-6=0

We add all the numbers together, and all the variables

-1u^2+5u-6=0

a = -1; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·(-1)·(-6)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*-1}=\frac{-6}{-2}
=+3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*-1}=\frac{-4}{-2}
=+2
$