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3r^2-2=67.12
We move all terms to the left:
3r^2-2-(67.12)=0
We add all the numbers together, and all the variables
3r^2-69.12=0
a = 3; b = 0; c = -69.12;
Δ = b2-4ac
Δ = 02-4·3·(-69.12)
Δ = 829.44
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{829.44}}{2*3}=\frac{0-\sqrt{829.44}}{6} =-\frac{\sqrt{}}{6} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{829.44}}{2*3}=\frac{0+\sqrt{829.44}}{6} =\frac{\sqrt{}}{6} $
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