Al-Khwarizmi's Completing-the-Square Construction

Draw a square of side $x$; its area is $x^2$, the first term of $x^2+bx=c$. Attach to two adjacent sides of this square two thin rectangles, each of width $b/2$ and length $x$, so their combined area is $2\cdot\tfrac b2\cdot x=bx$, the second term. The resulting L-shaped figure (a "gnomon") has area $x^2+bx=c$.

This gnomon is a big square of side $x+b/2$ with one small $\tfrac b2\times\tfrac b2$ corner square missing. To see this, notice the two rectangles plus the original square leave exactly a $(b/2)\times(b/2)$ gap at the outer corner where they meet. Fill that gap in: the whole figure — gnomon plus corner square — is now a genuine square of side $x+b/2$, whose area is therefore $c+(b/2)^2$ (the gnomon's area $c$ plus the corner square's area $(b/2)^2$, added exactly once).

So $(x+\tfrac b2)^2=c+\left(\tfrac b2\right)^2$. Taking the positive square root of both sides (side lengths are positive) gives $x+\dfrac b2=\sqrt{c+\left(\dfrac b2\right)^2}$, and isolating $x$ yields $x=\sqrt{c+\left(\dfrac b2\right)^2}-\dfrac b2$.

Algebraic check: expanding $\left(\sqrt{c+(b/2)^2}-b/2\right)^2+b\left(\sqrt{c+(b/2)^2}-b/2\right)$ term by term reduces, after the $\pm b\sqrt{c+(b/2)^2}\,b/2$ cross terms cancel, exactly to $c$ — confirming the formula independently of the picture. Numerically, for $x^2+10x=39$ (so $b=10,c=39$): $x=\sqrt{39+25}-5=\sqrt{64}-5=8-5=3$, and indeed $3^2+10\cdot3=9+30=39$.