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15t^2+19t-50=0
a = 15; b = 19; c = -50;
Δ = b2-4ac
Δ = 192-4·15·(-50)
Δ = 3361
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{3361}}{2*15}=\frac{-19-\sqrt{3361}}{30} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{3361}}{2*15}=\frac{-19+\sqrt{3361}}{30} $
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