Factor the tribunal into two binomials than solve?

Answer:
[tex](x-5)(x-3)=0[/tex]
[tex]x=5 \text{ and } x=3[/tex]
Step-by-step explanation:
Quadratic Equation given:
[tex]x^2-8x+15=0[/tex]
In order to factor this equation think that we'll have something in the format:[tex](x-a)(x-b)[/tex]
Once: -a -b = -c
and (-a)(-b)= d
Therefore, we have
[tex](x-5)(x-3)=0[/tex]
[tex]x-5=0\\\text{or}\\x-3=0[/tex]
[tex]x=5 \text{ and } x=3[/tex]