18=18b^-2*b^2

Solution for 18=18b^-2*b^2 equation:

18=18b^-2b^2

We move all terms to the left:

18-(18b^-2b^2)=0

We get rid of parentheses

2b^2-18b^+18=0

We add all the numbers together, and all the variables

2b^2-18b+18=0

a = 2; b = -18; c = +18;
Δ = b2-4ac
Δ = -182-4·2·18
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{5}}{2*2}=\frac{18-6\sqrt{5}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{5}}{2*2}=\frac{18+6\sqrt{5}}{4}
$