{maius segmentum
minus segmentum.

We translate the Latin passage, using the modern exponential notation and parentheses, as follows:

Given therefore an unequally divided line Z (10), and a rectangle beneath the segments AE (21) which is a gnomon. Half the difference of the segments ½X is given, and consequently the segment itself. For, if one of the two segments is placed equal to A, the other will be Z-A. Moreover, the rectangle is ZA-A²=AE. And because Z and AE are given, and there is ¼Z²-AE=¼X², and by 5c.18, ½Z+½X=A, and ½Z-½X=E, the equation will be solved thus: ½Z±√(¼Z²-AE)=A

{major segment
minor segment.

And so an equation having been proposed in which three species (terms) are in equally ascending powers, the highest species, moreover, being negative, the given magnitude which constitutes the middle species is the line to be bisected. And the given absolute magnitude to which it is equal is the rectangle beneath the unequal segments, without gnomon. As ZA-A²=AE, or in numbers, 10x-x²=21. And A or x is one of the two unequal segments. It may be found thus:

The half of the middle species is

Z
2
(5), its square is
Z²
4
(25). From it subtract the absolute term AE (21), and
Z²
4		-AE
(4) will be the square of half the difference of the segments. The square root of this,
√[(Z²
2		)²
-AE]
(2), is half the difference. If you add it to half the coefficient
Z
2
(5), or half the line to be bisected, the longer segment is obtained; if you subtract it, the smaller segment is obtained. I say:

Z
2		±√(Z²
4		-AE)=A {major segment
minor segment.

{major segment
minor segment.
Z
2
Z²
4
Z²
4		-AE
√[(Z²
2		)²
-AE>[
Z
2
Z
2		±√(Z²
4		-AE)=A{major segment
minor segment.