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4.905t^2+9.38t+-40.9=0
We add all the numbers together, and all the variables
4.905t^2+9.38t=0
a = 4.905; b = 9.38; c = 0;
Δ = b2-4ac
Δ = 9.382-4·4.905·0
Δ = 87.9844
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{-(9.38)-\sqrt{87.9844}}{2*4.905}=\frac{-9.38-\sqrt{87.9844}}{9.81} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9.38)+\sqrt{87.9844}}{2*4.905}=\frac{-9.38+\sqrt{87.9844}}{9.81} $
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