x2+(x-4)2=328

Solution for x2+(x-4)2=328 equation:

x2+(x-4)2=328

We move all terms to the left:

x2+(x-4)2-(328)=0

We add all the numbers together, and all the variables

x^2+(x-4)2-328=0

We multiply parentheses

x^2+2x-8-328=0

We add all the numbers together, and all the variables

x^2+2x-336=0

a = 1; b = 2; c = -336;
Δ = b2-4ac
Δ = 22-4·1·(-336)
Δ = 1348
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:

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