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6t^2-25t+26=0
a = 6; b = -25; c = +26;
Δ = b2-4ac
Δ = -252-4·6·26
Δ = 1
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{1}=1$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-1}{2*6}=\frac{24}{12} =2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+1}{2*6}=\frac{26}{12} =2+1/6 $
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