Part of the graph of a quadratic function is shown below. The graph passes through the point A (1, 15) and it cuts the x-axis at the points B and C. Given that C is (6, 0) and $x = 2$ is the line of symmetry of the graph,

1. Use the line of symmetry to write the vertex form of the quadratic.

Since the line of symmetry is $x=2$, the vertex has x-coordinate $2$.

So the equation is
$y=a(x-2)^2+k.$

2. Substitute the two given points to find $a$ and $k$.

Using $C(6,0)$:
$0=a(6-2)^2+k$
$16a+k=0 \quad \text{--- (1)}$

Using $A(1,15)$:
$15=a(1-2)^2+k$
$a+k=15 \quad \text{--- (2)}$

Subtract (2) from (1):
$15a=-15$
$a=-1$

Substitute into (2):
$-1+k=15$
$k=16$

Hence,
$y=-(x-2)^2+16.$

1. Substitute $y=4$ into the quadratic curve and solve for $x$.

From part (a),
$y=-(x-2)^2+16$
When $y=4$,
$4=-(x-2)^2+16$
$(x-2)^2=12$
$x-2=\pm\sqrt{12}=\pm2\sqrt{3}$
$x=2\pm2\sqrt{3}$
Since the positive value is required,
$x=2+2\sqrt{3}$