Answer
The solution set is {$-\frac{3}{2},-1$}.
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
$x=\frac{-5+1}{4}=-1$
$x=\frac{-5-1}{4}=\frac{-6}{4}=-\frac{3}{2}$
The solution set is {$-\frac{3}{2},-1$}.
