(x^2-26x)+169=4

Solution for (x^2-26x)+169=4 equation:

(x^2-26x)+169=4

We move all terms to the left:

(x^2-26x)+169-(4)=0

We add all the numbers together, and all the variables

(x^2-26x)+165=0

We get rid of parentheses

x^2-26x+165=0

a = 1; b = -26; c = +165;
Δ = b2-4ac
Δ = -262-4·1·165
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-4}{2*1}=\frac{22}{2}
=11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+4}{2*1}=\frac{30}{2}
=15
$