The sum of all possible values of x satisfying the equation $2^{4{\mathrm{x}}^2}-2^{2{\mathrm{x}}^2+\mathrm{x}+16}+2^{2\mathrm{x}+30}=0$ is

Consider the equation:- 

$2^{4{\mathrm{x}}^2}-2^{2{\mathrm{x}}^2+\mathrm{x}+16}+2^{2\mathrm{x}+30}=0$

⇒ $(2^{2{\mathrm{x}}^2}{)}^2-2^{2{\mathrm{x}}^2}{.2}^{\mathrm{x}+15}.2+(2^{\mathrm{x}+15}{)}^2=0$

⇒ $(2^{2{\mathrm{x}}^2}-2^{\mathrm{x}+15}{)}^2=0$

⇒ $2^{2{\mathrm{x}}^2}-2^{\mathrm{x}+15}=0$

⇒ $2^{2{\mathrm{x}}^2}=2^{\mathrm{x}+15}$

⇒ $2x^2=x+15$

⇒ $2x^2-x-15=0$

⇒ $2x^2-6x+5x-15=0$

⇒ $(2x+5)(x-3)=0$

Therefore, the possible values of x are $\frac{-5}{2}$ and 3 respectively. 

Hence, the sum of the possible values = $(3-\frac{5}{2})=\frac{1}{2}$