SOLUTION: if two vertices of an equilateral triangle are (-4,3) and(0,0), find the third vertex.

Here is one side of the equilateral triangle:
 

 
And as you can see there are two solutions. Let the third
vertex be (a,b):
 
and
 
First we find the distance from (0,0) to (-4,3) by the distance formula:
 
d =   
 
d = 
 
d = 
d = 
d =    
d = 5
 
So we know the distance from (0,0) to (a,b) must equal 5
so that the triangle will be equilateral.  So we use the
distance formula to make that equation:
 
 = 5
 = 5
Square both sides:
a² + b² = 25
 
We also know the distance from (-4,3) to (a,b) must also
equal 5 so that the triangle will be equilateral.  So we 
use the distance formula again to make that equation:
 
 = 5
 = 5
Square both sides:
           (a+4)² + (b-3)² = 25
a² + 8a + 16 + b² - 6b + 9 = 25
    a² + 8a + b² - 6b + 25 = 25
         a² + 8a + b² - 6b = 0
 
So we have the system of equations:
 
          a² + b² = 25         
a² + 8a + b² - 6b = 0
 
Subtracting the second equation from the first equation
gives the equation:
 
-8a + 6b = 25
      6b = 25 + 8a
       b = 
 
Substitute in
 
a² + b² = 25
 
a² +  = 25
a² +  = 25

Multiply thru by 36 to clear of fractions:

36a² + (25 + 8a)² = 900

36a² + 625 + 400a + 64a² = 900

100a² + 400a - 275 = 0

Divide thru by 25

4a² + 16a - 11 = 0

Use the quadratic formula:

a = 

a = 

a = 

a = 

a = 

a = 

a = 

Substituting in

6b = 25 + 8a

6b = 25 + 8()

6b = 25 + 4(-4 +- 3)

6b = 25 - 16 +- 12)

6b = 9 ± 12)

Divide through by 3

2b = 3 ± 4

b = 

So the third coordinate has two solutions:

(a,b) = (, )

 and

(a,b) = (, )

These are approximately:

(a,b) = (0.598, 4.964)  and

(a,b) = (-4.598, -1.964)

Edwin