Question

A problem that you will encounter in calculus 2 is finding the area under a curve. for the function f(x=-2x2-5x+3

Answer 1

$x$-intercepts occur at points $(x_i,0)$, so it follows that setting $f(x)=0$ and solving for $x$ will give you the $x$-coordinate of an intercept.

$-2x^2-5x+3=0\iff-(2x-1)(x+3)=0\implies x=\dfrac12,x=-3$

So the two intercepts are $(-3,0)$ and $\left(\dfrac12,0\right)$. For values of $x$ between the two intercepts, say at $x=0$, you have $f(0)=3>0$, which means the bounded region lies above the $x$-axis.

The area of the bounded region is then given explicitly by the definite integral

$\displaystyle\int_{-3}^{1/2}f(x)\,\mathrm dx$

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.