Answer
Visually, the parabola intersects the x-axis in two points: $x=-3$ and between $x=0$ and $x=1$. Let's assume that $x=\frac{1}{2}$. The x-intercepts are: $(-3,0)$ and $(\frac{1}{2},0)$
Algebraically: the x-intercepts are $(\frac{1}{2},0)$ and $(-3,0)$
Work Step by Step
$f(x)=0$
$2x^2+5x-3=0~~$ ($a=2,b=5,c=-3$):
$x=\frac{-5±\sqrt {5^2-4(2)(-3)}}{2(2)}=\frac{-5±\sqrt {49}}{4}=\frac{-5±7}{4}$
or $x=\frac{2}{4}=\frac{1}{2}$ or $x=-\frac{12}{4}=-3$
x-intercepts: $(\frac{1}{2},0)$ and $(-3,0)$
