## College Algebra (10th Edition)

$\displaystyle x=-\frac{1}{3}$
We solve by factoring: $(3x+4)^{2}-6(3x+4)+9=0$ $[(3x+4)-3][(3x+4)-3]=0$ $(3x+1)^2=0$ $3x+1=0$ $x=-\frac{1}{3}$