100=3.14(r)^2

Solution for 100=3.14(r)^2 equation:

100=3.14(r)^2

We move all terms to the left:

100-(3.14(r)^2)=0

We get rid of parentheses

-3.14r^2+100=0

a = -3.14; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-3.14)·100
Δ = 1256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1256}=\sqrt{4*314}=\sqrt{4}*\sqrt{314}=2\sqrt{314}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{314}}{2*-3.14}=\frac{0-2\sqrt{314}}{-6.28}
=-\frac{2\sqrt{314}}{-6.28}
=-\frac{\sqrt{314}}{-3.14}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{314}}{2*-3.14}=\frac{0+2\sqrt{314}}{-6.28}
=\frac{2\sqrt{314}}{-6.28}
=\frac{\sqrt{314}}{-3.14}
$