sqrt(x-1)=x

$\sqrt{x-1}=x$

$x-1=x^2$

$x^2-x+1=0$

$x=\frac{(-1)^2\pm\sqrt{4(1)(1)}}{2(1)}$<------ Quadratic Equation

$= \frac{1}{2}\pm\frac{2}{2}$

x = 3/2 or x=-1/2

There are two solutions.

You should check your results Max!  

You haven't done the quadratic solution correctly.

Oops. I was wrong

$\sqrt{x-1}=x$

$x-1=x^2$

$x^2-x+1=0$

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(1)}}{2(1)}$<------ Quadratic Equation

=$\frac{1}{2}\pm\frac{\sqrt{-3}}{2}$

=$\frac{1\pm\sqrt3i}{2}$

No real roots and 2 imaginary roots

Solve for x:
sqrt(x-1) = x

Raise both sides to the power of two:
x-1 = x^2

Subtract x^2 from both sides:
-x^2+x-1 = 0

Multiply both sides by -1:
x^2-x+1 = 0

Subtract 1 from both sides:
x^2-x = -1

Add 1/4 to both sides:
x^2-x+1/4 = -3/4

Write the left hand side as a square:
(x-1/2)^2 = -3/4

Take the square root of both sides:
x-1/2 = (i sqrt(3))/2 or x-1/2 = 1/2 (-i) sqrt(3)

Add 1/2 to both sides:
x = 1/2+(i sqrt(3))/2 or x-1/2 = 1/2 (-i) sqrt(3)

Add 1/2 to both sides:
Answer: |  x = 1/2+(i sqrt(3))/2     or      x = 1/2-(i sqrt(3))/2