$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$ - 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?

After the first line of your attempt, multiply both sides by $\sqrt{x+938^2}-\sqrt{x+140^2}$
$$x+938^2-(x+140^2)=1116(\sqrt{x+938^2}-\sqrt{x+140^2})$$
Divide both sides by $1116$.
$$\dfrac{938^2-140^2}{1116}=\sqrt{x+938^2}-\sqrt{x+140^2}$$
Taken with the previous equation
$$\sqrt{x+938^2}+\sqrt{x+140^2}=1116$$
we can add both equations to solve for $\sqrt{x+938^2}$ or subtract to solve for $\sqrt{x+140^2}$.  Then simply square both sides and solve for $x$.
