Babylonian quadratic equations

The Babylonians had an algorithm for solving quadratic equations of the form AX^2+BX=C, Their method only yields the positive answer, but since they had no negative numbers this wasn't much of a problem. To solve using their algorithm simply follow these steps,
1. reduce the equation so that A=1
2. multiply the value of B by 1/2
3. square the output of step 2
4. add the output of step 3 to value C
5. take the square root of the value from step 4
6. step 5 result-step 2 result equals the answer
Example: 2X^2+6X=12
1. x^2+3X=6
2. 3*1/2=1.5
3. 1.5^2=2.25
4. 6+2.25=8.25
5. square root of 8.25=2.87
6. 2.87-1.5=1.37
1.37 is the positive answer
To check your answer you can use the modern formula for finding the answers to quadratic equations. To use this formula, you take the equation in AX^2+BX=C form and simplify so A=1 and then subtract C from both sides so you get AX^2+BX-C=0, then useing the A, B, and C values from that equation simplify (-B± square root of (B^2-4AC))/2A to find x.

Babylonian cubic equations

The Babylonians could also solve cubic equations, though the answers are not as precise because they depended on a tablet with collections of values for one part of the proccess. For an equation of the form AX^3+bx^2=C the following method was used.
1. multiply both sides by A^2/B^3
2. were you have (A/B)*X substitute in N, you should get N^3+N^2=C
3. find the value of C on the chart below and use it to find the value of N
4. divide N by A/B to find the value of X

Table of values for cubic equation solving

Note that the values on the table are in Neugebauer Notation
Value of N Value of C
1 2
2 12
3 36
4 1,20
5 2,30
6 4,12
7 6,32
8 9,36
9 13,30
10 18,20
11 24,12
12 31,12
13 39,26
14 49,0
15 1,0
16 1,12,32
17 1,26,42
18 1,42,36
19 2,0,20
20 2,20,0

Exercises

Solve by Babylonian and modern methods,
3X^2+9X=36
2X^2+7X=42
X^2+35X=92
Solve by the Babylonian method,
3X^3+4X^2=20
X^3+2X^2=17
13X^3+7X^2=10
Translate the table from Neugegauer Notation to Hindu Arabic numerals.
Go to Pythagorean triples