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3r^2-2=50.92
We move all terms to the left:
3r^2-2-(50.92)=0
We add all the numbers together, and all the variables
3r^2-52.92=0
a = 3; b = 0; c = -52.92;
Δ = b2-4ac
Δ = 02-4·3·(-52.92)
Δ = 635.04
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{635.04}}{2*3}=\frac{0-\sqrt{635.04}}{6} =-\frac{\sqrt{}}{6} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{635.04}}{2*3}=\frac{0+\sqrt{635.04}}{6} =\frac{\sqrt{}}{6} $
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