-3t^2+6t=0

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Solution for -3t^2+6t=0 equation:



-3t^2+6t=0
a = -3; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-3)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-3}=\frac{-12}{-6} =+2 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-3}=\frac{0}{-6} =0 $

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