$\displaystyle 4*(5x+3)^2=9$

I am havning problems with the process to solve this Equation.

This is how far i have come:

$\displaystyle 4*(5x+3)^2=9$

$\displaystyle 4*(5x^2+30x+9)=9$ Using a rule here that i dont know the name of in english

$\displaystyle 20x^2+120x+36=9$

$\displaystyle 20x^2+120x=-27$

$\displaystyle 20x^2+x=-27/120$ now what? How do I get rid of the "^2" from 20x? or is this totaly wrong?

Fast answers are very much appreciated here!

Thank you in advance!

Rickard Liljeros