x2+(23-x)2=289

Solution for x2+(23-x)2=289 equation:

x2+(23-x)2=289

We move all terms to the left:

x2+(23-x)2-(289)=0

We add all the numbers together, and all the variables

x2+(-1x+23)2-289=0

We add all the numbers together, and all the variables

x^2+(-1x+23)2-289=0

We multiply parentheses

x^2-2x+46-289=0

We add all the numbers together, and all the variables

x^2-2x-243=0

a = 1; b = -2; c = -243;
Δ = b2-4ac
Δ = -22-4·1·(-243)
Δ = 976
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:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{61}}{2*1}=\frac{2-4\sqrt{61}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{61}}{2*1}=\frac{2+4\sqrt{61}}{2}
$