Find the range of values of $x$ that satisfy the inequality $5x - 3 < 2x(5 - x)$. The diagram shows a horizontal bridge of 100 metres supported by 2 pillars at the side, with a cable being suspended from point $A$ and $B$. Each pillar is vertical and is 40 metres tall. The lowest point $C$ of the cable is 10 meters above the bridge. The origin, $O$, is vertically below point $A$, where the foot of the pillar meets the bridge.
A quadratic function can be used to model the cable. Find the quadratic equation.

1. Expand the right hand side and rearrange the inequality into standard quadratic form.

$5x - 3 < 10x - 2x^2$
$2x^2 + 5x - 10x - 3 < 0$
$2x^2 - 5x - 3 < 0$

2. Factorise the quadratic expression to find the critical values.

$(2x + 1)(x - 3) < 0$

3. Determine the range of values for $x$ that satisfy the inequality.

The critical values are $x = -\frac{1}{2}$ and $x = 3$.
Since the quadratic curve $y = 2x^2 - 5x - 3$ opens upwards, the expression is negative between the roots.
$-\frac{1}{2} < x < 3$

1. Identify the coordinates of the vertex and the y-intercept from the given information

The origin $O$ is $(0,0)$. Point $A$ is at $(0, 40)$.
By symmetry, the lowest point $C$ (the vertex) is exactly halfway between the two pillars. The bridge is $100\text{ m}$ long, so the $x$-coordinate of $C$ is $50$.
The height of $C$ is $10\text{ m}$, so the vertex is $(50, 10)$.

2. Set up the quadratic equation using the vertex form and substitute the y-intercept to find the unknown constant

Let the equation be $y = a(x - h)^2 + k$.
Substituting the vertex $(50, 10)$:
$y = a(x - 50)^2 + 10$
Substitute the point $A(0, 40)$ into the equation:
$40 = a(0 - 50)^2 + 10$
$30 = 2500a$
$a = \frac{30}{2500} = \frac{3}{250} = 0.012$

3. Write the final quadratic equation

The quadratic equation is $y = 0.012(x - 50)^2 + 10$.
(Expanding gives $y = 0.012x^2 - 1.2x + 40$)