This is the question: $$\large{2^{2x^2}+2^{x^2 + 2x + 2} =2^{5+4x}}$$
What I did was put $~\large{2^{x^{2}}=t}$
From this, I got, roots of the quadratic:
$$\large{-2^{x+1}\pm~\left( 2^{2x+1}\sqrt{3}\right)}$$
Now this is equal to $\boxed{\large{2^{x^2}}}$. How do i find the value of x from this?
The given equation can be rewritten as $$2^{2x^2-4x+2}+2^{x^2-2x+4}=2^7$$Now take $2^{(x-1)^2}=z$ which gives you the equation $$z^2+8z=2^7\implies z=8,-16$$ Taking only the positive solution gives you, (assuming $x$ to be real)$$(x-1)^2=3\implies x=1\pm \sqrt{3}$$