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4.9t^2+10t-99=0
a = 4.9; b = 10; c = -99;
Δ = b2-4ac
Δ = 102-4·4.9·(-99)
Δ = 2040.4
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{-(10)-\sqrt{2040.4}}{2*4.9}=\frac{-10-\sqrt{2040.4}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{2040.4}}{2*4.9}=\frac{-10+\sqrt{2040.4}}{9.8} $
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