(2s+3)(3s+3)=0

Solution for (2s+3)(3s+3)=0 equation:

(2s+3)(3s+3)=0

We multiply parentheses ..

(+6s^2+6s+9s+9)=0

We get rid of parentheses

6s^2+6s+9s+9=0

We add all the numbers together, and all the variables

6s^2+15s+9=0

a = 6; b = 15; c = +9;
Δ = b2-4ac
Δ = 152-4·6·9
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3}{2*6}=\frac{-18}{12}
=-1+1/2
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3}{2*6}=\frac{-12}{12}
=-1
$