I'm calling it a Pseudo-Quadratic Equation?

We have: $\sqrt{17 + 8x - 2x^2} + \sqrt{4 + 12x - 3x^2}$ $=\sqrt{25 - 2(x-2)^2} + \sqrt{16 - 3(x-2)^2}$ $\leq (\sqrt{25} + \sqrt{16}) = 9 \leq \left(9 + (x-2)^2 \right) = (x^2 - 4x + 13)$ The equality throughout if and only if $x-2=0$ . Thus $x=\boxed{2}$ .

First, notice that if we let $a=17+8x-2x^2$ and $b = 4+12x-3x^2$ , we obtain $a-b=x^2-4x+13$ , which is precisely the RHS. So the equation becomes $\sqrt{a}+\sqrt{b}=a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$ and thus $\sqrt{a}-\sqrt{b}=1\implies \sqrt{17+8x-2x^2}=1+\sqrt{4+12x-3x^2}$ (Equation [1]) taking note the possibility $\sqrt{17+8x-2x^2}+\sqrt{4+12x-3x^2}=0$ (Equation [2]). We'll go back to that later.

Now, let $z = 4x - x^2$ . Then Equation [1] becomes

$\begin{aligned} \sqrt{17+8x-2x^2}&=1+\sqrt{4+12x-3x^2} \\ \implies \sqrt{17+2z}&=1+\sqrt{4+3z} \\ 17+2z&=5+3z+2\sqrt{4+3z} \\ 12-z&=2\sqrt{4+3z} \\ 144-24z+z^2&= 4(4+3z)=16+12z \\ z^2 - 36z + 128 = (z-4)(z-32) &= 0 \\ \implies z &= 4 \mathrm{\ or \ } z = 32\end{aligned}$

Upon checking on the equation $\sqrt{17+2z}=1+\sqrt{4+3z}$ , we find that $z=32$ is not an acceptable solution, but $z=4$ is. Then from our substitution, we get $z=4=4x-x^2 \implies x^2 - 4x + 4 = 0 \implies x = 2$

Now, we go back to equation [2] and solve for $x$ .

$\begin{aligned} \sqrt{17+8x-2x^2}+\sqrt{4+12x-3x^2}&=0 \\ \sqrt{17+8x-2x^2}&=-\sqrt{4+12x-3x^2} \\ 17+8x-2x^2 &= 4+12x-3x^2 \\ x^2 -4x + 13 &= 0 \end{aligned}$ which has no real solution. We therefore conclude that $x=2$ is the only real solution of the equation, and the answer is $\fbox{2}$ .

$\sqrt{17 + 8x - 2x^2} + \sqrt{4 + 12x - 3x^2} = x^2 - 4x + 13$

$\implies \sqrt{43 - 26 + 8x - 2x^2} + \sqrt{43 - 39 + 12x - 3x^2} = x^2 - 4x +13$

$\implies \sqrt{43 -2(x^2 - 4x + 13)} + \sqrt{43 -3(x^2 - 4x +13)} = x^2 - 4x + 13$

Take $u=x^2 - 4x +13$ , then the equation advances to:

$\implies \sqrt{43 -2u} + \sqrt{43 -3u} = u$

$\implies \sqrt{43 -2u} = u - \sqrt{43 -3u}$

Squaring both sides,

$\implies 43-2u = u^2 - 2u\sqrt{43-3u} + 43 - 3u$ , where we subtract $43$ from both the sides.

$\implies u^2 - u = 2u\sqrt{43-3u}$ , here we cancel out $u$ from both the sides concluding that $u=0$ is one of the solutions.

$\implies u - 1 = 2\sqrt{43-3u}$

Once again, squaring both sides we have:

$\implies u^2 - 2u + 1 = 4(43-3u) \implies u^2 - 2u + 1 = 172 - 12u \implies u^2 + 10u - 171 = 0 \implies \boxed{u=-19,9}$

Hence, we have $3$ values of $u$ to check for $(0,-19,9)$ .

Now, Case 1 : $u=0 \implies x^2 - 4x + 13 = 0 \implies x^2 - 4x + 4 = -9 \implies (x-2)^2 = -9 \implies x-2 = \pm 3i \implies x=2\pm3i$ .

So, it is not a real value.

Case 2

$u=-19 \implies x^2 - 4x + 13 = -19 \implies x^2 - 4x + 4 = -28 \implies (x-2)^2 = -28$ .

Evidently, we wont have any real value from this one.

Case 3

$u=9 \implies x^2 - 4x + 13 = 9 \implies x^2 - 4x + 4 = 0 \implies (x-2)^2 = 0 \implies \boxed{x=2}$

$2$ is real. This value when substituted into the parent equation satisfies itself. Therefore, there is only one real solution: $2$ .