Let A(3, 4), B(-2, 3) and C(x,y) be the three vertices of the equilateral triangle then, AB^{2}=BC^{2}=CA^{2}

Now, AB^{2}=BC^{2}

⇒ x^{2}+y^{2}+4x-6y+13=26

⇒ x^{2}+y^{2}+4x-6y-13=0 .......(i)

AB^{2}=CA^{2}

⇒ 26-(x^{2}+y^{2}-6x-8y+25)=0

⇒ x^{2}+y^{2}-6x-8y-1=0 ........(ii)

Subtracting (ii) from (i) we get,

10x+2y-12=0

⇒ 5x+y=6 ......(iii)

⇒ 5x=6-y

⇒ x=(6-y)/5

Subtracting x=(6-y)/5 in (i) we get

⇒ 26y^{2}-32y+6-325=0

⇒ 26y^{2}-32y-319=0

D=b^{2}-4ac

D=(-32)^{2}-4x26x(-319)

=1024+33176

=34200

Therefore, y=(32+185)/52=217/52 or y=(32-185)/52= -153/52

Substituting y=217/52 in (iii)

5x+217/52=6

5x=6-217/52=95/52

x=19/52

Again substituting y=-153/52 in (iii)

5x-153/52=6

5x=6+153/52=465/52

x=93/52

Therefore, the coordinates of the third vertex are (19/52,217/52) or (93/52,-153/52)