100(1+0.2P+0.02P^2)t=18

Solution for 100(1+0.2P+0.02P^2)t=18 equation:

100(1+0.2+0.02^2)P=18

We move all terms to the left:

100(1+0.2+0.02^2)P-(18)=0

We multiply parentheses

0P^2+100P+20P-18=0

We add all the numbers together, and all the variables

P^2+120P-18=0

a = 1; b = 120; c = -18;
Δ = b2-4ac
Δ = 1202-4·1·(-18)
Δ = 14472
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{14472}=\sqrt{36*402}=\sqrt{36}*\sqrt{402}=6\sqrt{402}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-6\sqrt{402}}{2*1}=\frac{-120-6\sqrt{402}}{2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+6\sqrt{402}}{2*1}=\frac{-120+6\sqrt{402}}{2}
$