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4.9t^2-28t-31=0
a = 4.9; b = -28; c = -31;
Δ = b2-4ac
Δ = -282-4·4.9·(-31)
Δ = 1391.6
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{-(-28)-\sqrt{1391.6}}{2*4.9}=\frac{28-\sqrt{1391.6}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+\sqrt{1391.6}}{2*4.9}=\frac{28+\sqrt{1391.6}}{9.8} $
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