b.3(b+6)=18

Solution for b.3(b+6)=18 equation:

b.3(b+6)=18

We move all terms to the left:

b.3(b+6)-(18)=0

We multiply parentheses

b^2+6b-18=0

a = 1; b = 6; c = -18;
Δ = b2-4ac
Δ = 62-4·1·(-18)
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{3}}{2*1}=\frac{-6-6\sqrt{3}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{3}}{2*1}=\frac{-6+6\sqrt{3}}{2}
$