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4.905t^2-6.946t+0.65=0
a = 4.905; b = -6.946; c = +0.65;
Δ = b2-4ac
Δ = -6.9462-4·4.905·0.65
Δ = 35.493916
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{-(-6.946)-\sqrt{35.493916}}{2*4.905}=\frac{6.946-\sqrt{35.493916}}{9.81} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6.946)+\sqrt{35.493916}}{2*4.905}=\frac{6.946+\sqrt{35.493916}}{9.81} $
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