For a quadratic equation of the form $$x^2+bx+c = 0,$$  we are able to factorise the left hand side expression into the form $$(x+p)(x+q),$$ if there exists a $p$ and $q$ such that $p+q=b$ and $p \times q = c$. 

For the expression $$\simplify{x^2+{b}x+{c}},$$

we need two numbers which add together to give $\var{b}$ and multiply together to give $\var{c}$. Therefore, $p=\var{p}$ and $q=\var{q}$, and the quadratic equation can be written in the form $$\simplify{(x+{p})(x+{q})} = 0.$$

Since one of the brackets must be $0$ for their product to be $0$ we know that
$$\simplify{(x+{p})}= 0 \quad \text{or} \quad \simplify{(x+{q})}= 0$$

and therefore, the solutions to this quadratic equation are 
$$x = \simplify{{-p}} \quad \text{and} \quad x = \simplify{{-q}} .$$