#### Answer

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

#### Work Step by Step

First, we should look at the discriminant ($b^2 - 4ac$) to determine how many real number solutions we have. The rules are such:
$(b^2 - 4ac) \gt 0$ then there are two real number solutions
$(b^2 - 4ac) \lt 0$ then there is one real number solution
$(b^2 - 4ac) = 0$ then there is no real number solution
Evaluating the discriminant:
Step 1: Write the equation in standard form.
$4x^2 - 4x +1 = 0$
Step 2: Determine our a, b, and c values
$a = 4, b = -4, c = 1$
Step 3: Evaluate $b^2 - 4ac$, using our values of a, b, and c
$b^2 - 4ac$ = $(-4)^2 - 4(4)(1) = 16 - 16 = 0$
Based on our rules of the discriminant, since $b^2 - 4ac = 0$, there is only one real number solutions.
Now, using the Quadratic Formula, we can find this real number solution. Step 1 and 2 are already complete from finding the discriminant.
Step 1: Write the equation in standard form.
$4x^2 - 4x +1 = 0$
Step 2: Determine our a, b, and c values
$a = 4, b = -4, c = 1$
Step 3: Substitute our values of a, b, and c into the formula
$x = \frac{-b \frac{+}{-} \sqrt(b^2 - 4ac)}{2a}$
$x = \frac{-(-4) \frac{+}{-} \sqrt((-4)^2 - 4(4)(1)}{2(4)}$
$x = \frac{4 \frac{+}{-} \sqrt(16 - 16)}{8}$
$x = \frac{4)}{8}$
$x = \frac{1}{2}$
We have found our real number solution.