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

Mathematics

2 answers:

GuDViN [60]3 years ago

8 0

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

cestrela7 [59]3 years ago

5 0

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

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

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