Chapter 9 - Solving Quadratic Equations - 9.5 - Solving Quadratic Equations Using the Quadratic Formula - Monitoring Progress - Page 518: 7

The equation has one real solution.

Comparing $-x^{2}+4x-4=0$ with $ax^{2}+bx+c$, we see that $a=-1, b=4$ and $c=-4$. $b^{2}-4ac=(4)^{2}-4(-1)(-4)=16-16=0$. The discriminant is $0$. So, the equation has one real solution.