Given the graph of y = 2x2 − x − 1, solve 2x2 − 3x − 3 = 0 and 2x2 − 2x − 1 = 0 using suitable straight lines.

Given graph: $y = 2x^2 - x - 1$
Equation to solve: $2x^2 - 3x - 3 = 0$

We want to express $2x^2 - 3x - 3 = 0$ in terms of $y = 2x^2 - x - 1$.
Rewrite $2x^2 - 3x - 3 = 0$ as:

Substitute $y$ for $(2x^2 - x - 1)$:

This is the equation of the suitable straight line.

To solve $2x^2 - 3x - 3 = 0$ graphically, you would draw the straight line $y = 2x + 2$ on the same axes as the graph of $y = 2x^2 - x - 1$. The solutions for $x$ are the x-coordinates of the points where the straight line intersects the parabola.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for $2x^2 - 3x - 3 = 0$ (where $a=2$, $b=-3$, $c=-3$):

The solutions are $x = \frac{3 + \sqrt{33}}{4}$ and $x = \frac{3 - \sqrt{33}}{4}$.
Approximately, $x \approx \frac{3 + 5.745}{4} \approx 2.186$ and $x \approx \frac{3 - 5.745}{4} \approx -0.686$.

The suitable straight line is $\boxed{y = 2x + 2}$.
The solutions are $\boxed{x = \frac{3 \pm \sqrt{33}}{4}}$.

c) Solve $2x^2 - 2x - 1 = 0$ using the graph of $y = 2x^2 - x - 1$.

Given graph: $y = 2x^2 - x - 1$
Equation to solve: $2x^2 - 2x - 1 = 0$