(3n+1)(4=n)

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

(3n+1)(4=n)

We move all terms to the left:

(3n+1)(4-(n))=0

We add all the numbers together, and all the variables

(3n+1)(-1n+4)=0

We multiply parentheses ..

(-3n^2+12n-1n+4)=0

We get rid of parentheses

-3n^2+12n-1n+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{-24}{-6}
=+4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*-3}=\frac{2}{-6}
=-1/3
$