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4.905t^2-120t-125=0
a = 4.905; b = -120; c = -125;
Δ = b2-4ac
Δ = -1202-4·4.905·(-125)
Δ = 16852.5
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{-(-120)-\sqrt{16852.5}}{2*4.905}=\frac{120-\sqrt{16852.5}}{9.81} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+\sqrt{16852.5}}{2*4.905}=\frac{120+\sqrt{16852.5}}{9.81} $
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