The Greeks thought of a product of two things as an area. For example, ab was thought of as a rectangle with base a and height b. This idea is used in Euclid's "proof" of the quadratic equation a(a  x) = x^{2} (or x^{2} + ax  a^{2} = 0). Note, Euclid's "proof" was more of a construction of the positive root of the equation, followed by a verification. Euclid's Proposition 11, from Book II of Elements is:
To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.
The goal is to find a point H on the line AB so that AB × HB equals AH × AH. Letting x = AH, and a = AB, then a  x = HB, and the equation to be solved for x is a(a  x) = x^{2} or x^{2} + ax  a^{2} = 0. AB, or a, is the given segment. From this, a square with side a is constructed. This being square ABDC in the figure above. AC is then bisected at point E. EB is drawn, and CA is extended to a point F so that EF = EB. Next draw square FGHA. Now H is the desired point so that x = AH is the positive root of x^{2} + ax  a^{2} = 0.
Below on the left is Euclid's verification. On the right modern notation and explanations are given.
By proposition II, 6  Prop. II, 6 is basically the identity  
CF × FG + AE^{2} = EF^{2}  (y + z)(y  z) + z^{2} = y^{2} 

(y + z)(y  z) = y^{2}  z^{2}  
where y = x + a/2 and z = a/2 so that  
By construction EF = EB;  y + z = x + a and y  z = x.  
thus  This gives us  
CF × FG + AE^{2} = EB^{2}  (a + x)(x) + (a/2)^{2} = (x + a/2)^{2} in (1) above.  
By Pythagorean Theorem,  By Pythagorean Theorem,  
CF × FG + AE^{2} = AB^{2} + AE^{2}  (a + x)(x) + (a/2)^{2} = a^{2} + (a/2)^{2}  
CF × FG 
(a + x)(x) 

AH^{2} = DB × HB  x^{2} = a(a  x)  
AH^{2} = AB × HB 
Thus H is the required point so that AH, or x, satisfies (2) on the left above.
You can try this by letting AB = a = 3 to get x^{2} + 3x  9 = 0. If you do the above construction, you should find that AX = x = 1.2426, which agrees with the positive root, from the quadratic formula.