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-4.9t^2+22.7t=0
a = -4.9; b = 22.7; c = 0;
Δ = b2-4ac
Δ = 22.72-4·(-4.9)·0
Δ = 515.29
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{-(22.7)-\sqrt{515.29}}{2*-4.9}=\frac{-22.7-\sqrt{515.29}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22.7)+\sqrt{515.29}}{2*-4.9}=\frac{-22.7+\sqrt{515.29}}{-9.8} $
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