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6t^2-19t-8=0
a = 6; b = -19; c = -8;
Δ = b2-4ac
Δ = -192-4·6·(-8)
Δ = 553
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{-(-19)-\sqrt{553}}{2*6}=\frac{19-\sqrt{553}}{12} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{553}}{2*6}=\frac{19+\sqrt{553}}{12} $
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