Solve the Quadratic Equation: \( \frac{x^2}{4} + \frac{2}{3}x + \frac{4}{9} = 0 \)

To solve the quadratic equation $\frac{x^2}{4}+\frac{2}{3}x+\frac{4}{9}=0$, we will use the quadratic formula:

The quadratic formula is given by:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, we identify the coefficients:
$a = \frac{1}{4}$, $b = \frac{2}{3}$, $c = \frac{4}{9}$

Now, we calculate the discriminant:
$b^2 - 4ac = \left(\frac{2}{3}\right)^2 - 4 \cdot \frac{1}{4} \cdot \frac{4}{9}$

Calculating further:
$b^2 = \frac{4}{9}$
$4ac = 4 \cdot \frac{1}{4} \cdot \frac{4}{9} = \frac{16}{36} = \frac{4}{9}$

Hence, the discriminant:
$b^2 - 4ac = \frac{4}{9} - \frac{4}{9} = 0$

Since the discriminant is 0, there is exactly one real solution. We apply the quadratic formula:
$x = \frac{-\left(\frac{2}{3}\right) \pm \sqrt{0}}{2 \times \frac{1}{4}}$

Simplify:
$x = \frac{-\frac{2}{3}}{\frac{1}{2}} = -\frac{2}{3} \cdot 2 = -\frac{4}{3}$

Therefore, the solution to the problem is $x = -\frac{4}{3}$.