The graphs of $y = a(x+2)(4-x)$ and $y = -\frac{1}{2}x + \frac{5}{2}$ are drawn on the grid.

1. Identify the x-intercepts from the factorised quadratic and find their midpoint.

For $y=a(x+2)(4-x)$, the x-intercepts are given by
$(x+2)(4-x)=0.$
So $x=-2$ or $x=4$.

The line of symmetry is halfway between the roots:
$x=\frac{-2+4}{2}=1.$

1. Relate the equation to intersections between the parabola and the horizontal line $y=k$, then use the range of the quadratic graph.

The equation $a(x+2)(4-x)=k$ represents the intersections of the curve $y=a(x+2)(4-x)$ with the horizontal line $y=k$.

Since $a=\frac12>0$ in $a(x+2)(4-x)$ and $(x+2)(4-x)$ has a negative coefficient of $x^2$, the curve is a downward-opening parabola. It has a maximum value, so horizontal lines above this maximum do not meet the curve. Therefore, for some values of $k$, there are no solutions.

1. Equate the two graph equations and substitute $a=\frac12$.

At the points of intersection,
$a(x+2)(4-x)=-\frac12x+\frac52.$
Using $a=\frac12$,
$\frac12(x+2)(4-x)=-\frac12x+\frac52.$

2. Clear fractions, expand and rearrange into $x^2+px+q=0$.

Multiply by $2$:
$(x+2)(4-x)=-x+5.$
Expand:
$-x^2+2x+8=-x+5.$
Rearrange:
$-x^2+3x+3=5$
$-x^2+3x-?$
Correctly collecting all terms:
$-x^2+2x+8+x-5=0$
$-x^2+3x+3=0.$
Multiply by $-1$:
$x^2-3x-3=0.$