4n^2-4=124

Solution for 4n^2-4=124 equation:

4n^2-4=124

We move all terms to the left:

4n^2-4-(124)=0

We add all the numbers together, and all the variables

4n^2-128=0

a = 4; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·4·(-128)
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{2}}{2*4}=\frac{0-32\sqrt{2}}{8}
=-\frac{32\sqrt{2}}{8}
=-4\sqrt{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{2}}{2*4}=\frac{0+32\sqrt{2}}{8}
=\frac{32\sqrt{2}}{8}
=4\sqrt{2}
$