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-3t^2+19t+42=0
a = -3; b = 19; c = +42;
Δ = b2-4ac
Δ = 192-4·(-3)·42
Δ = 865
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{865}}{2*-3}=\frac{-19-\sqrt{865}}{-6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{865}}{2*-3}=\frac{-19+\sqrt{865}}{-6} $
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