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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3u^2-4u+9u+12)=0

We get rid of parentheses

-3u^2-4u+9u+12=0

We add all the numbers together, and all the variables

-3u^2+5u+12=0

a = -3; b = 5; c = +12;
Δ = b2-4ac
Δ = 52-4·(-3)·12
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*-3}=\frac{-18}{-6}
=+3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*-3}=\frac{8}{-6}
=-1+1/3
$