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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4u^2-9u+12u+27)=0

We get rid of parentheses

-4u^2-9u+12u+27=0

We add all the numbers together, and all the variables

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

a = -4; b = 3; c = +27;
Δ = b2-4ac
Δ = 32-4·(-4)·27
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*-4}=\frac{-24}{-8}
=+3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*-4}=\frac{18}{-8}
=-2+1/4
$