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4.905t^2-14.79t-13=0
a = 4.905; b = -14.79; c = -13;
Δ = b2-4ac
Δ = -14.792-4·4.905·(-13)
Δ = 473.8041
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{-(-14.79)-\sqrt{473.8041}}{2*4.905}=\frac{14.79-\sqrt{473.8041}}{9.81} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14.79)+\sqrt{473.8041}}{2*4.905}=\frac{14.79+\sqrt{473.8041}}{9.81} $
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