n(n-1)+n(n-1)=168

Solution for n(n-1)+n(n-1)=168 equation:

n(n-1)+n(n-1)=168

We move all terms to the left:

n(n-1)+n(n-1)-(168)=0

We multiply parentheses

n^2+n^2-1n-1n-168=0

We add all the numbers together, and all the variables

2n^2-2n-168=0

a = 2; b = -2; c = -168;
Δ = b2-4ac
Δ = -22-4·2·(-168)
Δ = 1348
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:

$\sqrt{\Delta}=\sqrt{1348}=\sqrt{4*337}=\sqrt{4}*\sqrt{337}=2\sqrt{337}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{337}}{2*2}=\frac{2-2\sqrt{337}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{337}}{2*2}=\frac{2+2\sqrt{337}}{4}
$