40,000=3.14r^2

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Solution for 40,000=3.14r^2 equation:



40.000=3.14r^2
We move all terms to the left:
40.000-(3.14r^2)=0
We add all the numbers together, and all the variables
-(3.14r^2)+40=0
We get rid of parentheses
-3.14r^2+40=0
a = -3.14; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-3.14)·40
Δ = 502.4
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{502.4}}{2*-3.14}=\frac{0-\sqrt{502.4}}{-6.28} =-\frac{\sqrt{}}{-6.28} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{502.4}}{2*-3.14}=\frac{0+\sqrt{502.4}}{-6.28} =\frac{\sqrt{}}{-6.28} $

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