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4.9t^2-19.6t+14.7=0
a = 4.9; b = -19.6; c = +14.7;
Δ = b2-4ac
Δ = -19.62-4·4.9·14.7
Δ = 96.04
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{-(-19.6)-\sqrt{96.04}}{2*4.9}=\frac{19.6-\sqrt{96.04}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19.6)+\sqrt{96.04}}{2*4.9}=\frac{19.6+\sqrt{96.04}}{9.8} $
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