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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3u^2+13u+12=0

a = 3; b = 13; c = +12;
Δ = b2-4ac
Δ = 132-4·3·12
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*3}=\frac{-18}{6}
=-3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*3}=\frac{-8}{6}
=-1+1/3
$