We consider the functions $f(x)=-3x^2+x+4$ and $g(x)=5x-8$. We determine all $x$ such that $f(x)\leq g(x)$.
Step 1: Define h(x)
$h(x)=f(x)-g(x)=(-3x^2+x+4)-(5x-8)=-3x^2-4x+12$.
Step 2: Determine the zeros of h(x)
$D=(-4)^2-4\cdot-3\cdot 12 = 160$.
$x_1=\frac{4+\sqrt{128}}{2\cdot -3}=-\frac{2}{3}-\frac{2}{3}\sqrt{10}$, and
$x_2=\frac{4-\sqrt{128}}{2\cdot -3}=-\frac{2}{3}+\frac{2}{3}\sqrt{10}$.
Step 3: Make a sign chart
The following sign chart of $h(x)$ follows from $h(-10)=-248$, $h(0)=12$ and $h(5)=-83$.
Step 4: Observe the sign chart
If follows from the sign chart that for $x\leq -\frac{2}{3}-\frac{2}{3}\sqrt{10}$ and $x\geq -\frac{2}{3}+\frac{2}{3}\sqrt{10}$ it holds that $f(x) \leq g(x)$.
