How do I solve this radical equation? (the sqrt stands for square root)

$\displaystyle sqrt(x-3)+1=x$

$\displaystyle sqrt(x-3)=x-1$

$\displaystyle (sqrt(x-3))^2=(x-1)^2$

$\displaystyle x-3=(x-1)(x-1)$

$\displaystyle x-3=x^2-x-x-1$

$\displaystyle x-3=x^2-2x+1$

$\displaystyle 0=x^2-3x+4$

I need an actual number so I can check the solution.