(2s+3)(s+4)-228=0

Solution for (2s+3)(s+4)-228=0 equation:

(2s+3)(s+4)-228=0

We multiply parentheses ..

(+2s^2+8s+3s+12)-228=0

We get rid of parentheses

2s^2+8s+3s+12-228=0

We add all the numbers together, and all the variables

2s^2+11s-216=0

a = 2; b = 11; c = -216;
Δ = b2-4ac
Δ = 112-4·2·(-216)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-43}{2*2}=\frac{-54}{4}
=-13+1/2
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+43}{2*2}=\frac{32}{4}
=8
$