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6t^2+t-20=0
a = 6; b = 1; c = -20;
Δ = b2-4ac
Δ = 12-4·6·(-20)
Δ = 481
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{-(1)-\sqrt{481}}{2*6}=\frac{-1-\sqrt{481}}{12} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{481}}{2*6}=\frac{-1+\sqrt{481}}{12} $
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