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3m^2=0.2
We move all terms to the left:
3m^2-(0.2)=0
We add all the numbers together, and all the variables
3m^2-0.2=0
a = 3; b = 0; c = -0.2;
Δ = b2-4ac
Δ = 02-4·3·(-0.2)
Δ = 2.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{2.4}}{2*3}=\frac{0-\sqrt{2.4}}{6} =-\frac{\sqrt{}}{6} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{2.4}}{2*3}=\frac{0+\sqrt{2.4}}{6} =\frac{\sqrt{}}{6} $
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