Question

Ok update I already posted one but I have more work now. Still can't figure out how to finish it.

Answer 1

`\bf x^2+20x=476\implies \begin{array}{lcclll} x^2&+20x&-476=0\\ &\uparrow &\uparrow \\ &34-14&34\cdot -14 \end{array} \\\\\\ (x+34)(x-14)=0\implies x= \begin{cases} -34\\ 14 \end{cases}`

well, it can't be -34, since it's just an absolute unit, so is 14

Answer 2

you are basically at this point:
a^2+b^2=c^2
(10+x)^2=625-49
(10+x)^2=576

if you have not learned formulas to solve quadratic equations directly you can also "just" take the square root of both sides:
you have to remember that there are always two possiblites when finding a root: -a*-a=a^2 and a*a=a^2
(this is possibly your "factoring" hint)
so if you take the squareroot on both sides you get:
+/-(10+x)=+/-24

you know 24 has to be positive as it is the length of one side of the triangle
+/-(10+x)=24

2 solutions/equalities:
10+x=24
x=14

and
-10-x=24
-x=34
x=-34

obviously x+10>0 must hold true as your triangle side can only have a positive length, similar to 24 previously this leaves you only with x=14, because otherwise it would be 10-34<0

so x=14

This solution doesn't use the usual factoring as the hints seem to suggest, but instead for the squareroot
you might be supposed to find a slightly different way for the resolution in the end, but I didn't find it yet :)