CAT 2023 Slot 2 Solved Paper

It is given that $2^{4{\mathit{x}}^2}-2^{2{\mathit{x}}^2+\mathit{x}+16}+2^{2\mathit{x}+30}=0$, which can be written as:
$=>{\left(2^{2{\mathit{x}}^2}\right)}^2-2^{2{\mathit{x}}^2}\cdot 2^{\mathit{x}+15}\cdot 2^1+{\left(2^{\mathit{x}+15}\right)}^2=0$
$=>{\left(2^{2{\mathit{x}}^2}-2^{\mathit{x}+15}\right)}^2=0$
$=2^{2{\mathit{x}}^2}-2^{\mathit{x}+15}=0$ ( Since ${\left(\mathit{a}-\mathit{b}\right)}^2=0=>\mathit{a}-\mathit{b}=0)$
$=>2{\mathit{x}}^2=\mathit{x}+15$
$=2{\mathit{x}}^2-\mathit{x}-15=0$
$=>2{\mathit{x}}^2-6\mathit{x}+5\mathit{x}-15=0$
$\Rightarrow 2\mathit{x}\left(\mathit{x}-3\right)+5\left(\mathit{x}-3\right)=0$
$\Rightarrow \left(2\mathit{x}+5\right)\left(\mathit{x}-3\right)=0$
Hence, the possible values of $\mathit{x}$ are $-\frac{5}{2}$, and 3 , respectively.
Therefore, the sum of the possible values is $\left(3-\frac{5}{2}\right)=\frac{1}{2}$
The correct option is D