Answer
The equation has one real solution.
Work Step by Step
Given: $4x^{2}=4x-1$
$\implies 4x^{2}-4x+1=0$
Comparing $4x^{2}-4x+1=0$ with $ax^{2}+bx+c$, we see that
$a=4, b=-4$ and $c=1$.
$b^{2}-4ac=(-4)^{2}-4(4)(1)=0$.
The discriminant is equal to $0$. So, the equation has one real solution.
