We have $$ f(x) = 12\sqrt x $$ and $$ g(x) = x^2 - 7x + 12 $$
I need to find where they intersect. So far I've reduced the expression to $$ 12\sqrt x = x^2 - 7x + 12 $$ $$ 12 \sqrt x = (x-4)(x-3) $$ Square both sides and we get $$ 144x = (x-4)^2(x-3)^2 $$
Using a graph tools I get this visual and the points 0.51839, 9.62145 as a solution. But how do I get to these values myself?
Substitute $\sqrt{x}=t$, the equation converts to $$12t=t^4-7t^2+12$$ $$or \ t^4-7t^2-12t+12=0$$ Now this is what we call a depressed quartic equation and we can use method developed by Rene Descartes to solve such equations
let us say $$t^4-7t^2-12t+12=(t^2+at+b)(t^2+ct+d)=0$$ Then by expanding and comparing coefficients we have $$a+c=0, \ d+b+ac=-7$$ $$ad+bc = -12,\ bd = 12$$
We will have to use the cubic formula and we can solve for $a, b, c, d$ as $$a= -c=3.821844$$ $$b=5.373168,\ d=2.233322$$
And finally we can separately solve for $t^2+at+b=0$ and $t^2+ct+d=0$ using calculated values of $a, b, c, d$. The first one gives complex roots but for $t^2+ct+d=0$ putting $c=-3.821844$ and $d=2.233322$ we obtain two real roots as $t=3.10185$ and $0.719998$
Thus using $x=t^2$ we find $x$ to be equal to $0.518397$ and $9.621473$.
That's it.
EDIT: There is also a quartic formula for solving degree $4$ polynomials but just by looking at its form, it becomes obvious as why is it not very popular. For more info regarding quartic formula you may refer to https://en.wikipedia.org/wiki/Quartic_equation