4n^2+20=196

Solution for 4n^2+20=196 equation:

4n^2+20=196

We move all terms to the left:

4n^2+20-(196)=0

We add all the numbers together, and all the variables

4n^2-176=0

a = 4; b = 0; c = -176;
Δ = b2-4ac
Δ = 02-4·4·(-176)
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{11}}{2*4}=\frac{0-16\sqrt{11}}{8}
=-\frac{16\sqrt{11}}{8}
=-2\sqrt{11}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{11}}{2*4}=\frac{0+16\sqrt{11}}{8}
=\frac{16\sqrt{11}}{8}
=2\sqrt{11}
$