(3n+1)(n=4)

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Solution for (3n+1)(n=4) equation:



(3n+1)(n=4)
We move all terms to the left:
(3n+1)(n-(4))=0
We multiply parentheses ..
(+3n^2-12n+n-4)=0
We get rid of parentheses
3n^2-12n+n-4=0
We add all the numbers together, and all the variables
3n^2-11n-4=0
a = 3; b = -11; c = -4;
Δ = b2-4ac
Δ = -112-4·3·(-4)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*3}=\frac{-2}{6} =-1/3 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*3}=\frac{24}{6} =4 $

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