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6t^2-18t-108=0
a = 6; b = -18; c = -108;
Δ = b2-4ac
Δ = -182-4·6·(-108)
Δ = 2916
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2916}=54$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-54}{2*6}=\frac{-36}{12} =-3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+54}{2*6}=\frac{72}{12} =6 $
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