$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$ - I keep getting imaginary numbers $$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$$ 
My attempt
$\sqrt{x+938^2} + \sqrt{x + 140^2} = 1116$
$(\sqrt{x+938^2} + \sqrt{x + 140^2})^2 = (1116)^2$
$x+938^2  + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} + x + 140^2 = 1116^2$
$2x + 2*\sqrt{x+938^2}*\sqrt{x + 140^2}  = 1116^2 - 938^2 - 140^2$
$x + \sqrt{x^2 + 2(938^2 + 140^2)x+(938*140)^2} = 1116^2 - 938^2 - 140^2$
At this point trying to solve for x inside the sqrt in the quadratic gives me an imaginary number. How is it possible to solve this?

Repeated squaring is not necessary.  Let $y > -140$ be such that $x = y^2 + 280y$, so that $$\sqrt{x+140^2} - 140 = \sqrt{(y + 140)^2} - 140 = y.$$  Then $$\begin{align*} 38 &= \sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 \\ &= \sqrt{y^2 + 280y + 938^2} - 938 + y, \end{align*}$$ or equivalently, $$976-y = \sqrt{y^2 + 280y + 938^2}.$$  Now we square only once, giving $$y^2 - 2(976)y + 976^2 = y^2 + 280y + 938^2.$$  The quadratic terms cancel, and we get $$2(140+976)y = 976^2 - 938^2 = (976-938)(976+938) = 38(1914),$$ or $$y = \frac{6061}{186}.$$  Then substitute back to find the solution for $x$.
