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50.24=3.14(r^2)
We move all terms to the left:
50.24-(3.14(r^2))=0
We get rid of parentheses
-3.14r^2+50.24=0
a = -3.14; b = 0; c = +50.24;
Δ = b2-4ac
Δ = 02-4·(-3.14)·50.24
Δ = 631.0144
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{631.0144}}{2*-3.14}=\frac{0-\sqrt{631.0144}}{-6.28} =-\frac{\sqrt{}}{-6.28} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{631.0144}}{2*-3.14}=\frac{0+\sqrt{631.0144}}{-6.28} =\frac{\sqrt{}}{-6.28} $
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