40t-16t2=0

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Solution for 40t-16t2=0 equation:



40t-16t^2=0
a = -16; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·(-16)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*-16}=\frac{-80}{-32} =2+1/2 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*-16}=\frac{0}{-32} =0 $

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