F(x)=0.95x^2

Solution for F(x)=0.95x^2 equation:

(F)=0.95F^2

We move all terms to the left:

(F)-(0.95F^2)=0

We get rid of parentheses

-0.95F^2+F=0

a = -0.95; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-0.95)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-0.95}=\frac{-2}{-1.9}
=1+0.1/1.9
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-0.95}=\frac{0}{-1.9}
=0
$