50=3.14r^2

Solution for 50=3.14r^2 equation:

50=3.14r^2

We move all terms to the left:

50-(3.14r^2)=0

We get rid of parentheses

-3.14r^2+50=0

a = -3.14; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-3.14)·50
Δ = 628
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:

$\sqrt{\Delta}=\sqrt{628}=\sqrt{4*157}=\sqrt{4}*\sqrt{157}=2\sqrt{157}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{157}}{2*-3.14}=\frac{0-2\sqrt{157}}{-6.28}
=-\frac{2\sqrt{157}}{-6.28}
=-\frac{\sqrt{157}}{-3.14}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{157}}{2*-3.14}=\frac{0+2\sqrt{157}}{-6.28}
=\frac{2\sqrt{157}}{-6.28}
=\frac{\sqrt{157}}{-3.14}
$