[tex]x[/tex]-intercepts occur at points [tex](x_i,0)[/tex], so it follows that setting [tex]f(x)=0[/tex] and solving for [tex]x[/tex] will give you the [tex]x[/tex]-coordinate of an intercept.
[tex]-2x^2-5x+3=0\iff-(2x-1)(x+3)=0\implies x=\dfrac12,x=-3[/tex]
So the two intercepts are [tex](-3,0)[/tex] and [tex]\left(\dfrac12,0\right)[/tex]. For values of [tex]x[/tex] between the two intercepts, say at [tex]x=0[/tex], you have [tex]f(0)=3>0[/tex], which means the bounded region lies above the [tex]x[/tex]-axis.
The area of the bounded region is then given explicitly by the definite integral
[tex]\displaystyle\int_{-3}^{1/2}f(x)\,\mathrm dx[/tex]
Barring that (given your "no calc explanations allowed" caveat), the best you can do is to approximate the area of the region with basic shapes.
