$\left\{ \begin{aligned} 2^{x+2}=\frac{49}{4}x^2+4, \\ 2^{x+2}-4\le x^2(14-2^{x+2})*2^x \end{aligned} \right.$
Task is simple - solve this system.
Using some tricks I solved inequality $x \in (-\frac{2\sqrt3}{7}; \frac{2\sqrt3}{7})$.
But how to solve equation?
On
with a numerical method we find $$x=0$$ or $$0.246850313904907537017$$ or $$7.3972850014872596165$$ now you can findout which solutions lies in your interval
On
My idea is not completed way to find all roots but may be help the O.P to find some roots $$2^x=\frac{49}{16}x^2+1$$ $$x\log 2=\log(\frac{49}{16}x^2+1)$$ $$x=\frac{\log(\frac{49}{16}x^2+1)}{\log 2}$$ by using the iteration method, we can write the last equation as follow $$x_{n+1}=\frac{\log(\frac{49}{16}x_n^2+1)}{\log 2}$$ by using this equation we can get two roots only $$x_0=0$$ $$x_1=7.3972850014872596165$$
Well, as suspected, your edit brings a new piece of information (the inequation) that allows to easily solve that system using only high-school mathematics (the role of the inequality being to rule out the "ugly" non-zero solutions) - with the exception of a single step in the proof that seems to require very high-level mathematics.
For simplicity, let $t = 2^x$. The equation implies that $x^2 = \dfrac {16} {49} (t-1)$. Plugging this into the inequality and performing all the necessary simplifications brings it to
$$(t-1)(49 - 56t + 16t^2) \le 0 .$$
Inside the second pair of brackets one recognizes $(7 - 4t)^2$, so there are only two possibilities:
either $t-1 \ge 0$ and $(7 - 4t)^2 \le 0$,
or $t-1 \le 0$ and $(7 - 4t)^2 \ge 0$.
Since $(7 - 4t)^2 \ge 0$, the first possibility implies that $(7 - 4t)^2 = 0$, so that $t = \dfrac 7 4$, which also satisfies $t - 1 \ge 0$. Returning to $x$, this has the consequence that $2^x = \dfrac 7 4$ and $x^2 = \dfrac {12} {49}$, or equally well $x = \log_2 7 - 2$ and $x = \pm \dfrac {2 \sqrt 3} 7$. Since $x = \log_2 7 - 2 \ge \log_2 4 - 2 = 0$, it follows that $x$ cannot take that negative value. Can it take the positive one? Assuming it could, this would mean that $\log_2 7 = 2 + \dfrac {2 \sqrt 3} 7$, which after exponentiation would be equivalent to $\dfrac 7 4 = 4^ {\dfrac {\sqrt 3} 7}$. This, in turn, implies that $4^ {\dfrac {\sqrt 3} 7} \in \Bbb Q$, which after raising to the $7$th power implies that $4^ \sqrt 3 \in \Bbb Q$. This is not true, but I do not know how to prove this at a high-school level. Maybe you are supposed to use a scientific calculator to solve your problem? Anyway, that $4^ \sqrt 3$ is irrational is a consequence of the Gelfond-Schneider theorem, or of the Lindemann-Weierstrass theorem, both of them, though, being vastly out of the reach of high-school pupils. In any case, the conclusion so far is that $t = \dfrac 7 4$ does not lead to solutions of the given system.
The second possibility reduces to $t-1 \le 0$, because the square is always $\ge 0$, but since $x^2 = \dfrac {16} {49} (t-1)$, this would imply that $x^2 = 0$ ($0$ being the only number both $\ge 0$ and $\le 0$), which has the only solution $x=0$.
The analysis could have been performed equally well by replacing $t$ with $\dfrac {49} {16} x^2 + 1$ and performing all the compuations with $x$, instead of with $t$, but the reasoning would have reached the same difficult point about $\log_2 7$ that has been reached above.