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1.5t^2-24t=0
a = 1.5; b = -24; c = 0;
Δ = b2-4ac
Δ = -242-4·1.5·0
Δ = 576
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{576}=24$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24}{2*1.5}=\frac{0}{3} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24}{2*1.5}=\frac{48}{3} =16 $
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