3n^2-4n-4=0

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Solution for 3n^2-4n-4=0 equation: 



3n^2-4n-4=0

a = 3; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·3·(-4)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*3}=\frac{-4}{6}
=-2/3
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*3}=\frac{12}{6}
=2
$

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