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27=13r^2
We move all terms to the left:
27-(13r^2)=0
a = -13; b = 0; c = +27;
Δ = b2-4ac
Δ = 02-4·(-13)·27
Δ = 1404
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1404}=\sqrt{36*39}=\sqrt{36}*\sqrt{39}=6\sqrt{39}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{39}}{2*-13}=\frac{0-6\sqrt{39}}{-26} =-\frac{6\sqrt{39}}{-26} =-\frac{3\sqrt{39}}{-13} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{39}}{2*-13}=\frac{0+6\sqrt{39}}{-26} =\frac{6\sqrt{39}}{-26} =\frac{3\sqrt{39}}{-13} $
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