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200=3.14r^2
We move all terms to the left:
200-(3.14r^2)=0
We get rid of parentheses
-3.14r^2+200=0
a = -3.14; b = 0; c = +200;
Δ = b2-4ac
Δ = 02-4·(-3.14)·200
Δ = 2512
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2512}=\sqrt{16*157}=\sqrt{16}*\sqrt{157}=4\sqrt{157}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{157}}{2*-3.14}=\frac{0-4\sqrt{157}}{-6.28} =-\frac{4\sqrt{157}}{-6.28} =-\frac{2\sqrt{157}}{-3.14} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{157}}{2*-3.14}=\frac{0+4\sqrt{157}}{-6.28} =\frac{4\sqrt{157}}{-6.28} =\frac{2\sqrt{157}}{-3.14} $
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