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6t^2-2t=0
a = 6; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·6·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*6}=\frac{0}{12} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*6}=\frac{4}{12} =1/3 $
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