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100=3.14r(2)
We move all terms to the left:
100-(3.14r(2))=0
We add all the numbers together, and all the variables
-(+3.14r^2)+100=0
We get rid of parentheses
-3.14r^2+100=0
a = -3.14; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-3.14)·100
Δ = 1256
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1256}=\sqrt{4*314}=\sqrt{4}*\sqrt{314}=2\sqrt{314}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{314}}{2*-3.14}=\frac{0-2\sqrt{314}}{-6.28} =-\frac{2\sqrt{314}}{-6.28} =-\frac{\sqrt{314}}{-3.14} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{314}}{2*-3.14}=\frac{0+2\sqrt{314}}{-6.28} =\frac{2\sqrt{314}}{-6.28} =\frac{\sqrt{314}}{-3.14} $
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