The problem reads:

"If $a = -3$, $b = 4$, and $c = 10$, what is the value of the expression below?"

$\frac{2(b^2 - a^2)}{c} + \frac{1}{2}$

Step-by-step solution:

1. Substitute the values of $a$, $b$, and $c$:

   • $a = -3$
   • $b = 4$
   • $c = 10$

   So, the expression becomes:

   $\frac{2(4^2 - (-3)^2)}{10} + \frac{1}{2}$

2. Calculate $b^2$ and $a^2$:

   • $b^2 = 4^2 = 16$
   • $a^2 = (-3)^2 = 9$

   Now, substitute these values into the expression:

   $\frac{2(16 - 9)}{10} + \frac{1}{2}$

3. Simplify inside the parentheses:

   $16 - 9 = 7$

   Now, the expression becomes:

   $\frac{2 \times 7}{10} + \frac{1}{2}$

4. Multiply and simplify:

   $\frac{14}{10} + \frac{1}{2}$

5. Convert the fractions to have the same denominator:

   $\frac{14}{10} + \frac{5}{10} = \frac{14 + 5}{10} = \frac{19}{10}$

6. Simplify the result:

   $\frac{19}{10} = 1.9$

Thus, the value of the expression is 1.9.

Would you like further details or have any questions?

Here are 5 related questions for further practice:

1. What is the value of the expression $\frac{2(b^2 - a^2)}{c}$ if $a = 5$, $b = 2$, and $c = 3$?
2. How does the value of $a$ being negative affect the expression?
3. Can you solve a similar expression but with different constants, say $a = 1$, $b = 3$, $c = 8$?
4. What happens if $a = b$ in this type of expression?
5. How would you solve the expression if the second term was $\frac{1}{3}$ instead of $\frac{1}{2}$?

Tip: Always simplify powers and parentheses first to make calculations easier.