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(2*3.14r*r)+(2*3.14r*10)=440
We move all terms to the left:
(2*3.14r*r)+(2*3.14r*10)-(440)=0
We add all the numbers together, and all the variables
(+2*3.14r*r)+(+2*3.14r*10)-440=0
We get rid of parentheses
2*3.14r*r+2*3.14r*10-440=0
Wy multiply elements
6r^2+60r-440=0
a = 6; b = 60; c = -440;
Δ = b2-4ac
Δ = 602-4·6·(-440)
Δ = 14160
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{14160}=\sqrt{16*885}=\sqrt{16}*\sqrt{885}=4\sqrt{885}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-4\sqrt{885}}{2*6}=\frac{-60-4\sqrt{885}}{12} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+4\sqrt{885}}{2*6}=\frac{-60+4\sqrt{885}}{12} $
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