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-2.4t+0.24t^2=0
a = 0.24; b = -2.4; c = 0;
Δ = b2-4ac
Δ = -2.42-4·0.24·0
Δ = 5.76
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{-(-2.4)-\sqrt{5.76}}{2*0.24}=\frac{2.4-\sqrt{5.76}}{0.48} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.4)+\sqrt{5.76}}{2*0.24}=\frac{2.4+\sqrt{5.76}}{0.48} $
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