How to solve this system with equation and inequality? $\left\{
\begin{aligned}
2^{x+2}=\frac{49}{4}x^2+4, \\
2^{x+2}-4\le x^2(14-2^{x+2})*2^x
\end{aligned}
\right.$
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? 
 A: 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.
A: with a numerical method we find $$x=0$$ or $$0.246850313904907537017$$ or $$7.3972850014872596165$$ now you can findout which solutions lies in your interval
A: 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 
$$x_0=0$$
$$x_1=7.3972850014872596165$$
