Functional Quadratic

$f(x)$ 's intersection at $x=0$ is the $y$ -intercept, which is the constant term in the related equation. Since the vertex is always negative because the negative of an absolute value is always less than $0$ , and the $y$ -intercept is always positive since the square of any integer $c$ is positive, the y-intercept is greater than the vertex and the graph crosses the $x$ -axis. Since $c$ is an integer, the vertex cannot be $0$ as $-|2c^3 - 3|$ has an irrational solution, so $f(x)$ has $2$ distinct real solutions.

By factoring $6f(x)^2 + 7f(x) - 20=0$ into $(2f(x)+5)(3f(x)-4)=0$ , we can see that for some values $d$ and $b$ , $f(d) = -2.5$ and $f(b) = 1.333$ . If $f(x) - 1.333 = 0$ , $1.333 < c^2 + 2$ so the $y$ -intercept will be positive. Because this is a downwards translation, the vertex will still be negative. As a result, $f(x) = 1.333$ has $2$ distinct real solutions.

When we consider the other equation, $f(x) + 2.5 = 0$ , the y-intercept will be positive as $c^2 + 2 + 2.5 > 0$ . However, if $c=1$ , the vertex will be $-1$ , which is greater than $-2.5$ . Since adding $2.5$ is a vertical translation upwards, $-1+2.5 = 1.5$ so the vertex will become positive. Because the vertex and $y$ -intercept are positive, $f(x) = -2.5$ has no solutions.

We determined $2$ solutions for $f(x) = 1.333$ and $0$ solutions for $f(x) = -2.5$ , so in total, there are $0 + 2 = 2$ real distinct solutions.