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81+r^2=3r^2
We move all terms to the left:
81+r^2-(3r^2)=0
We add all the numbers together, and all the variables
-2r^2+81=0
a = -2; b = 0; c = +81;
Δ = b2-4ac
Δ = 02-4·(-2)·81
Δ = 648
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{2}}{2*-2}=\frac{0-18\sqrt{2}}{-4} =-\frac{18\sqrt{2}}{-4} =-\frac{9\sqrt{2}}{-2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{2}}{2*-2}=\frac{0+18\sqrt{2}}{-4} =\frac{18\sqrt{2}}{-4} =\frac{9\sqrt{2}}{-2} $
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