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4.905t^2+1t-10=0
We add all the numbers together, and all the variables
4.905t^2+t-10=0
a = 4.905; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·4.905·(-10)
Δ = 197.2
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{-(1)-\sqrt{197.2}}{2*4.905}=\frac{-1-\sqrt{197.2}}{9.81} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{197.2}}{2*4.905}=\frac{-1+\sqrt{197.2}}{9.81} $
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