#### Answer

Solution set: $\{4,10\}$

#### Work Step by Step

$g(a)=7$
$\displaystyle \frac{a}{2}+\frac{20}{a}=7$
Exclude solutions which yield 0 in denominators:
$a\neq 0\qquad (*)$
Multiply the equation with the LCD=$2a$
$a^{2}+40=14a$
$a^{2}-14a+40=0$
... $(x^{2}+bx+c$ is factored by searching for factors of c whose sum is b).
... -10 and -4 are factors of 40 whose sum is -14
$(a-10)(a-4)=0$
By the zero product principle,
$a\in\{4,10\}, $ ... both values satisfy (*).
Solution set: $\{4,10\}$