Solving Quadratic Equations by Factorising

The equation $(x+3)(x+4)=0$ can be solved by setting each bracket equal to zero. This gives $x+3=0$ and $x+4=0$, leading to $x=-3$ and $x=-4$.

The equation $(x-2)(x+9)=0$ can be solved by setting each bracket equal to zero. This gives $x-2=0$ and $x+9=0$, leading to $x=2$ and $x=-9$.

To factorise $x^2+7x+10=0$, we need to find two numbers that multiply to give 10 and add up to 7. These numbers are 2 and 5, so the factored form is $(x+2)(x+5)=0$.

To factorise $x^2-11x+28=0$, we need to find two numbers that multiply to give 28 and add up to -11. These numbers are -4 and -7, so the factored form is $(x-4)(x-7)=0$.

To factorise $x^2-3x-18=0$, we need to find two numbers that multiply to -18 and add up to -3. These numbers are -6 and 3, so the factored form is $(x-6)(x+3)=0$.

First divide the equation by 5 to simplify it to $x^2+6x-16=0$. Then, find two numbers that multiply to -16 and add to 6. These numbers are -2 and 8, so the factorised form is $(x-2)(x+8)=0$. Therefore, the solutions are $x=2$ and $x=-8$.