3.14r^2=10

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Solution for 3.14r^2=10 equation:



3.14r^2=10
We move all terms to the left:
3.14r^2-(10)=0
a = 3.14; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·3.14·(-10)
Δ = 125.6
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{125.6}}{2*3.14}=\frac{0-\sqrt{125.6}}{6.28} =-\frac{\sqrt{}}{6.28} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{125.6}}{2*3.14}=\frac{0+\sqrt{125.6}}{6.28} =\frac{\sqrt{}}{6.28} $

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