(3n+n)(n+1)-126=0

Solution for (3n+n)(n+1)-126=0 equation:

(3n+n)(n+1)-126=0

We add all the numbers together, and all the variables

(+4n)(n+1)-126=0

We multiply parentheses ..

(+4n^2+4n)-126=0

We get rid of parentheses

4n^2+4n-126=0

a = 4; b = 4; c = -126;
Δ = b2-4ac
Δ = 42-4·4·(-126)
Δ = 2032
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2032}=\sqrt{16*127}=\sqrt{16}*\sqrt{127}=4\sqrt{127}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{127}}{2*4}=\frac{-4-4\sqrt{127}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{127}}{2*4}=\frac{-4+4\sqrt{127}}{8}
$