3.14r^2=20

Solution for 3.14r^2=20 equation:

3.14r^2=20

We move all terms to the left:

3.14r^2-(20)=0

a = 3.14; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·3.14·(-20)
Δ = 251.2
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{251.2}}{2*3.14}=\frac{0-\sqrt{251.2}}{6.28}
=-\frac{\sqrt{}}{6.28}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{251.2}}{2*3.14}=\frac{0+\sqrt{251.2}}{6.28}
=\frac{\sqrt{}}{6.28}
$