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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2u^2+u+6=0

a = -2; b = 1; c = +6;
Δ = b2-4ac
Δ = 12-4·(-2)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-2}=\frac{-8}{-4}
=+2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-2}=\frac{6}{-4}
=-1+1/2
$