Comment reoudre -21x-4xau carre =0

[tex]Bonjour;\\\\\\-21x-4x^2=0\ ;\\\\\\donc\ :\ 21x+4x^2=0\ ;\\\\\\donc\ :\ 4x(\dfrac{21}{4}+x)=0\ ;\\\\\\donc\ :\ x=0\ ou\ \dfrac{21}{4}+x=0\ ;\\\\\\donc\ :\ x=0\ ou\ x=-\dfrac{21}{4}\ .[/tex]