The graph of $y = x^2 - 3x - 1.75$ is shown on the grids below.

1. Identify the coefficients of the quadratic and apply the axis of symmetry formula.

For $y=x^2-3x-1.75$, $a=1$ and $b=-3$.

Line of symmetry:
$x=-\frac{b}{2a}=-\frac{-3}{2(1)}=\frac{3}{2}=1.5$

1. Rearrange the given equation to make $y$ the subject.

$4y=7x-16$
$y=\frac{7x-16}{4}=\frac{7}{4}x-4$

2. Choose two convenient points on the line, then draw the straight line on the given grid.

Using $y=\dfrac{7}{4}x-4$:

When $x=0$, $y=-4$, so one point is $(0,-4)$.

When $x=4$, $y=3$, so another point is $(4,3)$.

Plot $(0,-4)$ and $(4,3)$, then join them with a straight line for $-1\le x\le 5$.

1. Rewrite the quadratic equation so it can be related to the parabola and the line on the graph.

Given
$4x^2-19x+9=0,$
rewrite as
$4x^2-12x-7=7x-16.$
Since the parabola is $y=x^2-3x-1.75$,
$4y=4x^2-12x-7.$
So the equation is solved where
$4y=7x-16,$
i.e. where the line intersects the parabola.

2. Read the $x$-coordinates of the intersection points from the graph.

From the graph, the line $4y=7x-16$ intersects the parabola $y=x^2-3x-1.75$ at approximately
$x=0.5 \quad \text{and} \quad x=4.3.$