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3m^2+3m-1=0
a = 3; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·3·(-1)
Δ = 21
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{-(3)-\sqrt{21}}{2*3}=\frac{-3-\sqrt{21}}{6} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{21}}{2*3}=\frac{-3+\sqrt{21}}{6} $
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