(r+3)(3r+4)=0

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

(r+3)(3r+4)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3r^2+13r+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:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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