-6t^2+24t=0

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Solution for -6t^2+24t=0 equation:



-6t^2+24t=0
a = -6; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·(-6)·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*-6}=\frac{-48}{-12} =+4 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*-6}=\frac{0}{-12} =0 $

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