(3s+11)(s+1)=-4

Solution for (3s+11)(s+1)=-4 equation:

(3s+11)(s+1)=-4

We move all terms to the left:

(3s+11)(s+1)-(-4)=0

We add all the numbers together, and all the variables

(3s+11)(s+1)+4=0

We multiply parentheses ..

(+3s^2+3s+11s+11)+4=0

We get rid of parentheses

3s^2+3s+11s+11+4=0

We add all the numbers together, and all the variables

3s^2+14s+15=0

a = 3; b = 14; c = +15;
Δ = b2-4ac
Δ = 142-4·3·15
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*3}=\frac{-18}{6}
=-3
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*3}=\frac{-10}{6}
=-1+2/3
$