4^2+x^2=300^2

Solution for 4^2+x^2=300^2 equation:

4^2+x^2=300^2

We move all terms to the left:

4^2+x^2-(300^2)=0

We add all the numbers together, and all the variables

x^2-89984=0

a = 1; b = 0; c = -89984;
Δ = b2-4ac
Δ = 02-4·1·(-89984)
Δ = 359936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{359936}=\sqrt{256*1406}=\sqrt{256}*\sqrt{1406}=16\sqrt{1406}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{1406}}{2*1}=\frac{0-16\sqrt{1406}}{2}
=-\frac{16\sqrt{1406}}{2}
=-8\sqrt{1406}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{1406}}{2*1}=\frac{0+16\sqrt{1406}}{2}
=\frac{16\sqrt{1406}}{2}
=8\sqrt{1406}
$