(4u+3)(9-u)=0

Solution for (4u+3)(9-u)=0 equation:

(4u+3)(9-u)=0

We add all the numbers together, and all the variables

(4u+3)(-1u+9)=0

We multiply parentheses ..

(-4u^2+36u-3u+27)=0

We get rid of parentheses

-4u^2+36u-3u+27=0

We add all the numbers together, and all the variables

-4u^2+33u+27=0

a = -4; b = 33; c = +27;
Δ = b2-4ac
Δ = 332-4·(-4)·27
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-39}{2*-4}=\frac{-72}{-8}
=+9
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+39}{2*-4}=\frac{6}{-8}
=-3/4
$