=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)=0


We calculate terms in parentheses: -(100(1+0.2+0.02^2)P), so:

100(1+0.2+0.02^2)P

We multiply parentheses

0P^2+100P+20P

We add all the numbers together, and all the variables

P^2+120P

Back to the equation:

-(P^2+120P)


We get rid of parentheses

-P^2-120P=0

We add all the numbers together, and all the variables

-1P^2-120P=0

a = -1; b = -120; c = 0;
Δ = b2-4ac
Δ = -1202-4·(-1)·0
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-120}{2*-1}=\frac{0}{-2}
=0
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+120}{2*-1}=\frac{240}{-2}
=-120
$