500=3.14r^2

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Solution for 500=3.14r^2 equation:



500=3.14r^2
We move all terms to the left:
500-(3.14r^2)=0
We get rid of parentheses
-3.14r^2+500=0
a = -3.14; b = 0; c = +500;
Δ = b2-4ac
Δ = 02-4·(-3.14)·500
Δ = 6280
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:
$\sqrt{\Delta}=\sqrt{6280}=\sqrt{4*1570}=\sqrt{4}*\sqrt{1570}=2\sqrt{1570}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1570}}{2*-3.14}=\frac{0-2\sqrt{1570}}{-6.28} =-\frac{2\sqrt{1570}}{-6.28} =-\frac{\sqrt{1570}}{-3.14} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1570}}{2*-3.14}=\frac{0+2\sqrt{1570}}{-6.28} =\frac{2\sqrt{1570}}{-6.28} =\frac{\sqrt{1570}}{-3.14} $

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