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-4.9t^2+29t-14=0
a = -4.9; b = 29; c = -14;
Δ = b2-4ac
Δ = 292-4·(-4.9)·(-14)
Δ = 566.6
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{-(29)-\sqrt{566.6}}{2*-4.9}=\frac{-29-\sqrt{566.6}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{566.6}}{2*-4.9}=\frac{-29+\sqrt{566.6}}{-9.8} $
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