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4.905t^2+12t=0
a = 4.905; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·4.905·0
Δ = 144
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*4.905}=\frac{-24}{9.81} =-2+3.1739130434783/7.1086956521739 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*4.905}=\frac{0}{9.81} =0 $
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