F(5)=3.14r2

Solution for F(5)=3.14r2 equation:

(5)=3.14F^2

We move all terms to the left:

(5)-(3.14F^2)=0

We get rid of parentheses

-3.14F^2+5=0

a = -3.14; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-3.14)·5
Δ = 62.8
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{62.8}}{2*-3.14}=\frac{0-\sqrt{62.8}}{-6.28}
=-\frac{\sqrt{}}{-6.28}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{62.8}}{2*-3.14}=\frac{0+\sqrt{62.8}}{-6.28}
=\frac{\sqrt{}}{-6.28}
$