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4.905t^2+18t=-4
We move all terms to the left:
4.905t^2+18t-(-4)=0
We add all the numbers together, and all the variables
4.905t^2+18t+4=0
a = 4.905; b = 18; c = +4;
Δ = b2-4ac
Δ = 182-4·4.905·4
Δ = 245.52
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{-(18)-\sqrt{245.52}}{2*4.905}=\frac{-18-\sqrt{245.52}}{9.81} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+\sqrt{245.52}}{2*4.905}=\frac{-18+\sqrt{245.52}}{9.81} $
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