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4.905t^2+40t-16.5=0
a = 4.905; b = 40; c = -16.5;
Δ = b2-4ac
Δ = 402-4·4.905·(-16.5)
Δ = 1923.73
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{-(40)-\sqrt{1923.73}}{2*4.905}=\frac{-40-\sqrt{1923.73}}{9.81} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+\sqrt{1923.73}}{2*4.905}=\frac{-40+\sqrt{1923.73}}{9.81} $
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