## Algebra 2 Common Core

$$-3,3$$
A quadratic equation of the form $$x^{2}+bx+c$$ can be factored as $$=(x+x_{1})(x+x_{2}),$$ where $x_{1}+x_{2}=b$ and $x_{1}\times x_{2}=c.$ To solve $$(x^{2}-9)=0,$$ factor $$(x^{2}-9)=(x+3)(x-3)=0.$$ Hence, the solutions are $$x=-3, 3.$$