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6t^2-5t-11=0
a = 6; b = -5; c = -11;
Δ = b2-4ac
Δ = -52-4·6·(-11)
Δ = 289
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{289}=17$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-17}{2*6}=\frac{-12}{12} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+17}{2*6}=\frac{22}{12} =1+5/6 $
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