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

Sagot :

[tex]{ x }^{ 2 }+9x+20=0\\ \\ \left( x+5 \right) \left( x+4 \right) =0\\ \\ \therefore \quad x=-5,\quad x=-4.[/tex]
konrad509
Ok, let assume it's [tex]x^2[/tex]

[tex]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[/tex]
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