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((2r)^2)-r^2=48
We move all terms to the left:
((2r)^2)-r^2-(48)=0
determiningTheFunctionDomain -r^2+2r^2-48=0
We add all the numbers together, and all the variables
r^2-48=0
a = 1; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·1·(-48)
Δ = 192
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*1}=\frac{0-8\sqrt{3}}{2} =-\frac{8\sqrt{3}}{2} =-4\sqrt{3} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*1}=\frac{0+8\sqrt{3}}{2} =\frac{8\sqrt{3}}{2} =4\sqrt{3} $
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