F(3)=3.14r^2

Solution for F(3)=3.14r^2 equation:

(3)=3.14F^2

We move all terms to the left:

(3)-(3.14F^2)=0

We get rid of parentheses

-3.14F^2+3=0

a = -3.14; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-3.14)·3
Δ = 37.68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{37.68}}{2*-3.14}=\frac{0-\sqrt{37.68}}{-6.28}
=-\frac{\sqrt{}}{-6.28}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{37.68}}{2*-3.14}=\frac{0+\sqrt{37.68}}{-6.28}
=\frac{\sqrt{}}{-6.28}
$