This seems familiar.

Rewrite the equation as $\sqrt { 10 } { ({ 5 }^{ x }) }^{ 2 }-7\times ({ 5 }^{ x })\times ({ 2 }^{ x })+\sqrt { 10 } { ({ 2 }^{ x }) }^{ 2 }=0$ . This is a quadratic equation in ${ 5 }^{ x }$ . Solve for ${ 5 }^{ x }$ . ${ 5 }^{ x }=\frac { 7\times { 2 }^{ x }\quad \pm \quad \sqrt { 49\times { 2 }^{ 2x }-40\times { 2 }^{ 2x } } }{ 2\sqrt { 10 } } \quad =\quad \frac { 7\times { 2 }^{ x }\quad \pm \quad 3\times { 2 }^{ x } }{ 2\sqrt { 10 } }$ . Solutions are ${ 5 }^{ x }=\frac { { 2 }^{ x }\sqrt { 5 } }{ \sqrt { 2 } } \quad and\quad { 5 }^{ x }=\frac { { 2 }^{ x }\times \sqrt { 2 } }{ \sqrt { 5 } } i.e\quad \quad x=\cfrac { 1 }{ 2 } ,-\cfrac { 1 }{ 2 }$ . Hence Sum of Roots= 0