4t2+12t=0

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Solution for 4t2+12t=0 equation:



4t^2+12t=0
a = 4; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·4·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*4}=\frac{-24}{8} =-3 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*4}=\frac{0}{8} =0 $

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