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1000=3.14r^2
We move all terms to the left:
1000-(3.14r^2)=0
We get rid of parentheses
-3.14r^2+1000=0
a = -3.14; b = 0; c = +1000;
Δ = b2-4ac
Δ = 02-4·(-3.14)·1000
Δ = 12560
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12560}=\sqrt{16*785}=\sqrt{16}*\sqrt{785}=4\sqrt{785}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{785}}{2*-3.14}=\frac{0-4\sqrt{785}}{-6.28} =-\frac{4\sqrt{785}}{-6.28} =-\frac{2\sqrt{785}}{-3.14} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{785}}{2*-3.14}=\frac{0+4\sqrt{785}}{-6.28} =\frac{4\sqrt{785}}{-6.28} =\frac{2\sqrt{785}}{-3.14} $
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