By using factorisation, solve $(x - 3)(2x + 8) = -12. Answer: x = ... Solve the equation $\frac{4}{2x-3} - \frac{3}{x+2} = 1.$ Give your solutions correct to two decimal places. Answer: x = ...

1. Expand the left-hand side and rearrange into quadratic form

$(x-3)(2x+8)=-12$
$2x^2+8x-6x-24=-12$
$2x^2+2x-24=-12$
$2x^2+2x-12=0$

2. Simplify the quadratic equation and factorise

$2x^2+2x-12=0$
Divide by $2$:
$x^2+x-6=0$
$(x+3)(x-2)=0$

3. Use the zero product property to find the values of $x$

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

1. State the restrictions and multiply by the common denominator

$2x-3 \ne 0$ and $x+2 \ne 0$, so $x \ne \dfrac32$ and $x \ne -2$.

Multiply both sides of
$\frac{4}{2x-3}-\frac{3}{x+2}=1$
by $(2x-3)(x+2)$:
$4(x+2)-3(2x-3)=(2x-3)(x+2).$

2. Expand and rearrange into quadratic form

$4x+8-6x+9=2x^2+x-6$
$-2x+17=2x^2+x-6$
$2x^2+3x-23=0$

3. Solve the quadratic equation and round to two decimal places

Using $a=2$, $b=3$, $c=-23$,
$x=\frac{-3\pm\sqrt{3^2-4(2)(-23)}}{2(2)}$
$x=\frac{-3\pm\sqrt{193}}{4}$
$x=2.723\ldots \text{ or } x=-4.223\ldots$
Correct to two decimal places,
$x=2.72 \text{ or } x=-4.22.$