(3u+1)(u+27)=0

Solution for (3u+1)(u+27)=0 equation:

(3u+1)(u+27)=0

We multiply parentheses ..

(+3u^2+81u+u+27)=0

We get rid of parentheses

3u^2+81u+u+27=0

We add all the numbers together, and all the variables

3u^2+82u+27=0

a = 3; b = 82; c = +27;
Δ = b2-4ac
Δ = 822-4·3·27
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(82)-80}{2*3}=\frac{-162}{6}
=-27
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(82)+80}{2*3}=\frac{-2}{6}
=-1/3
$