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6t^2+3.6t=0
a = 6; b = 3.6; c = 0;
Δ = b2-4ac
Δ = 3.62-4·6·0
Δ = 12.96
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{-(3.6)-\sqrt{12.96}}{2*6}=\frac{-3.6-\sqrt{12.96}}{12} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.6)+\sqrt{12.96}}{2*6}=\frac{-3.6+\sqrt{12.96}}{12} $
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