Please solve the quadratic inequality shown in the picture.

Let's solve the quadratic inequality $$2x^2 - 3x + 2 > 0$$.

First, we need to find the discriminant of the quadratic equation $$2x^2 - 3x + 2 = 0$$.

The discriminant ($$D$$) is given by the formula: $$D = b^2 - 4ac$$ where $$a = 2$$, $$b = -3$$, and $$c = 2$$.

Plugging in the values, we get: $$D = (-3)^2 - 4(2)(2) = 9 - 16 = -7$$

Since the discriminant $$D$$ is negative ($$D < 0$$), the quadratic equation $$2x^2 - 3x + 2 = 0$$ has no real roots. This means the parabola $$y = 2x^2 - 3x + 2$$ does not intersect the x-axis.

Now, we need to determine if the parabola opens upwards or downwards. Since the coefficient of $$x^2$$ (which is $$a$$) is positive ($$a = 2 > 0$$), the parabola opens upwards.

Since the parabola opens upwards and does not intersect the x-axis, the inequality $$2x^2 - 3x + 2 > 0$$ is true for all real values of $$x$$. In other words, the parabola is always above the x-axis.

Therefore, the solution to the inequality $$2x^2 - 3x + 2 > 0$$ is all real numbers.

Answer: $$x \in (-\infty, +\infty)$$