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24.5t-4.9t^2=0
a = -4.9; b = 24.5; c = 0;
Δ = b2-4ac
Δ = 24.52-4·(-4.9)·0
Δ = 600.25
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{-(24.5)-\sqrt{600.25}}{2*-4.9}=\frac{-24.5-\sqrt{600.25}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24.5)+\sqrt{600.25}}{2*-4.9}=\frac{-24.5+\sqrt{600.25}}{-9.8} $
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