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

Solution for (2u-3)(1-u)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -2; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·(-2)·(-3)
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*-2}=\frac{-6}{-4}
=1+1/2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*-2}=\frac{-4}{-4}
=1
$