Euclid's geometric solution of a quadratic equation
We find x so that (a−x)x=b2.
Draw a line of length a.
Draw a line of length b perpendicular to the line a from its midpoint.
Draw a circle centred at the end of the line with radius 21a.
Now x is a solution of (a−x)x=b2.
To see this use the fact that (21a)2−(21a−x)2=b2.
But (21a)2−(21a−x)2=(a−x)x so x is the required solution.