Answer
The solution set is {$0,\frac{9}{4}$}.
Work Step by Step
$4x^2-9x=0$
$a=4, b=-9, c=0$
$D=b^2-4ac=(-9)^2-4\cdot4(0)=81-0=81>0$ The equation has two real solutions.
$x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{9\pm\sqrt 81}{2\cdot4}$ Use the quadratic formula.
$x=\frac{9\pm9}{8}$
$x=\frac{18}{8}$ , $x=\frac{9-9}{8}$
$x=\frac{9}{4}$,$x=0$
The solution set is {$0,\frac{9}{4}$}.