The Latex bragger strikes again!
https://latex.artofproblemsolving.com/miscpdf/ujuyolds.pdf?t=1566795355878 We can find all the solutions of the equation $|x^2-14x+29|=4$ by splitting it up into two cases: $$x^2-14x+29=4$$ $$-x^2+14x-29=4$$ Looking at the first case, where $x^2-14x+29\geq 0$, we begin to notice that in the form of a quadratic $ax^2+bx+c$ $a$ is 1 and $b$ is even, making the scenario the easiest to complete the square. $$x^2-14x+29=4$$ $$x^2-14x=-33$$ $$x^2-14x+49=-33+49$$ $$(x-7)^2=16$$ $$x-7=\pm 4$$ We can then split this into two cases again: $$x-7=+4$$ $$x-7=-4$$ $$x=11$$ $$x=3$$ Now that we found the values of $x$ when $x^2-14x+29\geq 0$, we can move on to our second case of when $x^2-14x+29<0$ $$-x^2+14x-29=4$$ $$-x^2+14x-33=0$$ $$(-x+11)(x-3)=0$$ $$x=11, 3$$ Interesting... Now that we have our final two values of $3, 11$ we have to check these values in case there are many erroneous cases since in the first case we took the square root of both sides. ...