(4u-3)(2-u)=0

Solution for (4u-3)(2-u)=0 equation:

(4u-3)(2-u)=0

We add all the numbers together, and all the variables

(4u-3)(-1u+2)=0

We multiply parentheses ..

(-4u^2+8u+3u-6)=0

We get rid of parentheses

-4u^2+8u+3u-6=0

We add all the numbers together, and all the variables

-4u^2+11u-6=0

a = -4; b = 11; c = -6;
Δ = b2-4ac
Δ = 112-4·(-4)·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*-4}=\frac{-16}{-8}
=+2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*-4}=\frac{-6}{-8}
=3/4
$