Answer
The solution set is {$1,\frac{3}{2}$}.
Work Step by Step
$2x^2-5x+3=0$
$a=2, b=-5, c=3$
$D=b^2-4ac=(-5)^2-4\cdot2(3)=25-24=1>0$ The equation has two real solutions.
$x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{5\pm\sqrt 1}{2\cdot2}$ Use the quadratic formula.
$x=\frac{5\pm\sqrt 1}{4}$ Simplify
$x=\frac{5+1}{4}=\frac{6}{4}=\frac{3}{2}$
$x=\frac{5-1}{4}=1$
The solution set is {$1,\frac{3}{2}$}.
