(2s+3)(s+4)=0

Solution for (2s+3)(s+4)=0 equation:

(2s+3)(s+4)=0

We multiply parentheses ..

(+2s^2+8s+3s+12)=0

We get rid of parentheses

2s^2+8s+3s+12=0

We add all the numbers together, and all the variables

2s^2+11s+12=0

a = 2; b = 11; c = +12;
Δ = b2-4ac
Δ = 112-4·2·12
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*2}=\frac{-16}{4}
=-4
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*2}=\frac{-6}{4}
=-1+1/2
$