P(x)=100/0.02x^2+1

Solution for P(x)=100/0.02x^2+1 equation:

(P)=100/0.02P^2+1

We move all terms to the left:

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


Domain of the equation: 0.02P^2+1)!=0

P∈R


We get rid of parentheses

P-100/0.02P^2-1=0

We multiply all the terms by the denominator


P*0.02P^2-1*0.02P^2-100=0

Wy multiply elements

0P^2+0P-100=0

We add all the numbers together, and all the variables

P^2+P-100=0

a = 1; b = 1; c = -100;
Δ = b2-4ac
Δ = 12-4·1·(-100)
Δ = 401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{401}}{2*1}=\frac{-1-\sqrt{401}}{2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{401}}{2*1}=\frac{-1+\sqrt{401}}{2}
$