(2s+3)(s+4)=288

Solution for (2s+3)(s+4)=288 equation:

(2s+3)(s+4)=288

We move all terms to the left:

(2s+3)(s+4)-(288)=0

We multiply parentheses ..

(+2s^2+8s+3s+12)-288=0

We get rid of parentheses

2s^2+8s+3s+12-288=0

We add all the numbers together, and all the variables

2s^2+11s-276=0

a = 2; b = 11; c = -276;
Δ = b2-4ac
Δ = 112-4·2·(-276)
Δ = 2329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{2329}}{2*2}=\frac{-11-\sqrt{2329}}{4}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{2329}}{2*2}=\frac{-11+\sqrt{2329}}{4}
$