Question

I,m trying to solve this algebra question: x^+9x+20=0

Answer 1

${ x }^{ 2 }+9x+20=0\\ \\ \left( x+5 \right) \left( x+4 \right) =0\\ \\ \therefore \quad x=-5,\quad x=-4.$

Answer 2

Ok, let assume it's $x^2$

$x^2+9x+20=0\\ x^2+4x+5x+20=0\\ x(x+4)+5(x+4)=0\\ (x+5)(x+4)=0\\ x=-5 \vee x=-4$