40=3.1416r^2

Solution for 40=3.1416r^2 equation:

40=3.1416r^2

We move all terms to the left:

40-(3.1416r^2)=0

We get rid of parentheses

-3.1416r^2+40=0

a = -3.1416; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-3.1416)·40
Δ = 502.656
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{502.656}}{2*-3.1416}=\frac{0-\sqrt{502.656}}{-6.2832}
=-\frac{\sqrt{}}{-6.2832}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{502.656}}{2*-3.1416}=\frac{0+\sqrt{502.656}}{-6.2832}
=\frac{\sqrt{}}{-6.2832}
$