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14t+4.9t^2=0
a = 4.9; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·4.9·0
Δ = 196
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{196}=14$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*4.9}=\frac{-28}{9.8} =-2+6/7 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*4.9}=\frac{0}{9.8} =0 $
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