College Algebra 7th Edition

$(\frac{2}{3},+\infty)$
$3x^{2}-5x+2$ is a real number when $3x^{2}-5x+2\geq0$ We have $3x^{2}-5x+2=(x-1)(x-\frac{2}{3})$ is zero at $x=\frac{2}{3}, x=1$ $x\lt\frac{2}{3}$ is negative, $3x^{2}-5x+2$ is not defined as a real number.