## College Algebra (6th Edition)

$x=0,\frac{1}{3}$
Distribute by multiplying on the left side. Then put all terms on the left side to make the other side zero. Then find the GCF of the terms and set both resulting terms equal to zero to solve for $x$. $2x(x-3)=5x^2-7x$ $2x^2-6x=5x^2-7x$ $-3x^2+x=0$ $x(-3x+1)=0$ $x=0$ $-3x+1=0$ $-3x=-1$ $x=\frac{1}{3}$ $x=0,\frac{1}{3}$