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(2t^2)+13t-60=0
a = 2; b = 13; c = -60;
Δ = b2-4ac
Δ = 132-4·2·(-60)
Δ = 649
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{-(13)-\sqrt{649}}{2*2}=\frac{-13-\sqrt{649}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{649}}{2*2}=\frac{-13+\sqrt{649}}{4} $
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