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4.9t^2+12t-2.5=0
a = 4.9; b = 12; c = -2.5;
Δ = b2-4ac
Δ = 122-4·4.9·(-2.5)
Δ = 193
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{-(12)-\sqrt{193}}{2*4.9}=\frac{-12-\sqrt{193}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+\sqrt{193}}{2*4.9}=\frac{-12+\sqrt{193}}{9.8} $
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