3r^2+12r=0

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Solution for 3r^2+12r=0 equation: 



3r^2+12r=0

a = 3; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·3·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*3}=\frac{-24}{6}
=-4
$


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

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