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4.9t^2+13t-25=0
a = 4.9; b = 13; c = -25;
Δ = b2-4ac
Δ = 132-4·4.9·(-25)
Δ = 659
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{659}}{2*4.9}=\frac{-13-\sqrt{659}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{659}}{2*4.9}=\frac{-13+\sqrt{659}}{9.8} $
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