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4-4r^2=2r^2-2
We move all terms to the left:
4-4r^2-(2r^2-2)=0
We get rid of parentheses
-4r^2-2r^2+2+4=0
We add all the numbers together, and all the variables
-6r^2+6=0
a = -6; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-6)·6
Δ = 144
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-6}=\frac{-12}{-12} =1 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-6}=\frac{12}{-12} =-1 $
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