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3r^2=6
We move all terms to the left:
3r^2-(6)=0
a = 3; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·3·(-6)
Δ = 72
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*3}=\frac{0-6\sqrt{2}}{6} =-\frac{6\sqrt{2}}{6} =-\sqrt{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*3}=\frac{0+6\sqrt{2}}{6} =\frac{6\sqrt{2}}{6} =\sqrt{2} $
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