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1.68t^2+6t-142.2=0
a = 1.68; b = 6; c = -142.2;
Δ = b2-4ac
Δ = 62-4·1.68·(-142.2)
Δ = 991.584
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{-(6)-\sqrt{991.584}}{2*1.68}=\frac{-6-\sqrt{991.584}}{3.36} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+\sqrt{991.584}}{2*1.68}=\frac{-6+\sqrt{991.584}}{3.36} $
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