<s>[INST] How to solve for $ x$ when $x^2 \times  a = (x-10)^2 \times b$ I'm trying to figure this one out, and I'm stuck. I'm going:  
$x^2 \times a = (x-10)^2 \times b $ 
$x^2 \times a/b = (x-10)^2 /b $ 
$a/b = \frac{(x-10)^2} { x^2} /  x^2$  
$\sqrt{a/b} = (x-10)/x$     
I'm not sure how I proceed or if this is even the best way to start. (I do fine with everything up to linear algebra.)
 [/INST] $\textbf{Hint}$ : Try to put it into the form $Dx^2 + Ex + F$ and use the quadratic formula to find the solutions. Move cursor over the box for more details.

 $ax^2 = b(x - 10)^2$
 $ax^2 = b(x^2 - 20x + 100)$
 $ax^2 = bx^2 - 20bx + 100b$
 $0 = (b - a)x^2 - 20bx + 100b$
 Now using the quadratic formula :
 $x = \frac{20b \pm \sqrt{(20b)^2 - 4(100)b(b - a)}}{2(b - a)}$
 $= \frac{20b \pm \sqrt{400ab}}{2(b-a)} = \frac{20b \pm 20\sqrt{ab}}{2(b-a)}$
 $= 10 \frac{b \pm \sqrt{ab}}{b - a}$

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