x²+5x-27=x+5=y², so x²+5x-x-27-5=0, x²+4x-32=(x+8)(x-4)=0, and x=-8 or 4.
y²=x+5=-8+5=-3 or y²=4+5=9, so y=±3. Two real solutions appear to be (x,y)=(4,3) and (4,-3), but we need to test them first:
x=4: y²=x²+5x-27⇒9=16+20-27=9 (true) for y=-3 or 3. There is no real solution for y when x=-8, so there are only two real solutions for (x,y).