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5+13t-4.9t^2=0
a = -4.9; b = 13; c = +5;
Δ = b2-4ac
Δ = 132-4·(-4.9)·5
Δ = 267
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{-(13)-\sqrt{267}}{2*-4.9}=\frac{-13-\sqrt{267}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{267}}{2*-4.9}=\frac{-13+\sqrt{267}}{-9.8} $
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