Using factorisation, solve $9x^2 + 6x - 8 = 0$. Answer x = ... or ... Hence solve $9(y-1)^2 + 6y - 14 = 0$. Answer y = ... or ...

1. Factorise the quadratic expression.

$9x^2+6x-8=0$
Find two factors whose product is $9\times(-8)=-72$ and whose sum is $6$: $12$ and $-6$.
$9x^2+12x-6x-8=0$
$3x(3x+4)-2(3x+4)=0$
$(3x+4)(3x-2)=0$

2. Set each factor equal to zero and solve for $x$.

$(3x+4)(3x-2)=0$
So,
$3x+4=0 \quad \text{or} \quad 3x-2=0$
$x=-\dfrac{4}{3} \quad \text{or} \quad x=\dfrac{2}{3}$

1. Rewrite the equation in terms of $y-1$ to match the quadratic solved in part (a).

Since $6y-14=6(y-1)-8$,
$9(y-1)^2+6y-14=0$
becomes
$9(y-1)^2+6(y-1)-8=0.$

2. Use the roots from part (a) and solve for $y$.

From part (a), for $9x^2+6x-8=0$,
$x=\frac{2}{3}\quad\text{or}\quad x=-\frac{4}{3}.$
Here, $x=y-1$.

So
$y-1=\frac{2}{3}\quad\text{or}\quad y-1=-\frac{4}{3}.$
Hence
$y=\frac{5}{3}\quad\text{or}\quad y=-\frac{1}{3}.$