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19.62t^2+19t-58=0
a = 19.62; b = 19; c = -58;
Δ = b2-4ac
Δ = 192-4·19.62·(-58)
Δ = 4912.84
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{-(19)-\sqrt{4912.84}}{2*19.62}=\frac{-19-\sqrt{4912.84}}{39.24} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{4912.84}}{2*19.62}=\frac{-19+\sqrt{4912.84}}{39.24} $
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