6b+b(b+2)=108

Solution for 6b+b(b+2)=108 equation:

6b+b(b+2)=108

We move all terms to the left:

6b+b(b+2)-(108)=0

We multiply parentheses

b^2+6b+2b-108=0

We add all the numbers together, and all the variables

b^2+8b-108=0

a = 1; b = 8; c = -108;
Δ = b2-4ac
Δ = 82-4·1·(-108)
Δ = 496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{31}}{2*1}=\frac{-8-4\sqrt{31}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{31}}{2*1}=\frac{-8+4\sqrt{31}}{2}
$