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24t-16t^2=0
a = -16; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·(-16)·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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*-16}=\frac{-48}{-32} =1+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*-16}=\frac{0}{-32} =0 $
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