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18t^2-17t+1=0
a = 18; b = -17; c = +1;
Δ = b2-4ac
Δ = -172-4·18·1
Δ = 217
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-\sqrt{217}}{2*18}=\frac{17-\sqrt{217}}{36} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{217}}{2*18}=\frac{17+\sqrt{217}}{36} $
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