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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-2u^2-3u+8u+12)=0

We get rid of parentheses

-2u^2-3u+8u+12=0

We add all the numbers together, and all the variables

-2u^2+5u+12=0

a = -2; b = 5; c = +12;
Δ = b2-4ac
Δ = 52-4·(-2)·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*-2}=\frac{-16}{-4}
=+4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*-2}=\frac{6}{-4}
=-1+1/2
$