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6t^2-13t+6=0
a = 6; b = -13; c = +6;
Δ = b2-4ac
Δ = -132-4·6·6
Δ = 25
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{25}=5$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5}{2*6}=\frac{8}{12} =2/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*6}=\frac{18}{12} =1+1/2 $
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