12t^2-4t=0

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Solution for 12t^2-4t=0 equation: 



12t^2-4t=0

a = 12; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·12·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*12}=\frac{0}{24}
=0
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*12}=\frac{8}{24}
=1/3
$

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