Estate E and Estate W are 35 km apart. Jason cycles at an average speed of $x$ km/h from Estate E to Estate W.
On his return journey, he cycles 4 km/h faster.

1. Express the time taken using distance divided by speed.

Distance from Estate E to Estate W $=35$ km.

Speed from Estate E to Estate W $=x$ km/h.

$\text{Time}=\frac{\text{distance}}{\text{speed}}=\frac{35}{x}$

1. Express the return journey time using distance divided by the faster speed.

Return speed $=x+4$ km/h.

$\text{Time}=\frac{35}{x+4}\text{ h}$

1. Convert the time difference into hours.

$20$ minutes $=\dfrac{20}{60}=\dfrac{1}{3}$ h

2. Form an equation using the fact that the return journey is faster.

Time from Estate E to Estate W is $\dfrac{35}{x}$ h.

Time from Estate W to Estate E is $\dfrac{35}{x+4}$ h.

Since the return journey is $\dfrac{1}{3}$ h faster,
$\dfrac{35}{x}-\dfrac{35}{x+4}=\dfrac{1}{3}.$

3. Clear the fractions and simplify to the required quadratic equation.

$\dfrac{35}{x}-\dfrac{35}{x+4}=\dfrac{1}{3}$

Multiply throughout by $3x(x+4)$:
$105(x+4)-105x=x(x+4).$

$105x+420-105x=x^2+4x$

$420=x^2+4x$

$x^2+4x-420=0.$

1. Factorise the quadratic expression.

Find two numbers with product $-420$ and sum $4$: $21$ and $-20$.

$x^2+4x-420=(x+22)(x-18)$

So
$x^2+4x-420=0$
becomes
$(x+22)(x-18)=0.$

2. Use the zero product property to form two linear equations.

$(x+22)(x-18)=0$

Hence
$x+22=0 \quad \text{or} \quad x-18=0.$

3. Solve the two resulting linear equations.

$x+22=0 \Rightarrow x=-22$

$x-18=0 \Rightarrow x=18$

1. Choose the physically valid speed and find the outward journey time.

From part (c), $x=18$ or $x=-22$. Since speed cannot be negative, $x=18$.

Time from Estate E to Estate W:
$\frac{35}{x}=\frac{35}{18}\text{ h}$