\({ x }^{ 2 }-10x+9\)
First you should ask, which two numbers add up to -10 and multiply to 9.
-9 and -1 add up to -10 and multiply to 9.
\((-9)+(-1)=-10\)
\((-9)(-1)=9\)
So we can factor and rewrite like this ;
\((x-9)\cdot (x-1)\)
Let's equal it to 0 again.
\(\\ (x-9)\cdot (x-1)=0\)
The multiplication of (x-9) and (x-1) is equal to 0
So we should equal them to 0 seperately to find the solution set
\(x-9=0\\ \\ x=0+9\\ \\ x=9\)
and
\(x-1=0\\ \\ x=0+1\\ \\ x=1\)
The solution set,
\({ x }_{ 1,2 }\quad =\quad \left\{ 9,\quad 1 \right\}\)
