You may already be familiar with this question type (solving equations with variables in the exponents), but there's something a little different about this question.
With most other questions, the base is a constant (i.e., NOT a variable).
For example: \(5^{2x+1} = 5^{x-3}\)
Here, we can automatically conclude that \(2x+1 = x-3\), which we can then solve for \(x\)
This conclusion is based on the following property: If \(b^x = b^y\), then \(x = y\)
IMPORTANT: The above property has 3 provisos: \(b \neq 0, b \neq 1\) and \(b \neq -1\)

These provisos should make sense.
For example, \(1^x\) always equals 1. So, if an equation has \(1\) in the base, we really can't make any conclusions about the exponents.
So, if \(1^x = 1^3\), we can't then conclude that \(x=3\), since ANY value of \(x\) will satisfy the equation:
In other words, since \(1^x = 1^3=1^8=1^2=1^1=1^{-2} = 1...\) etc, we can't make any conclusions about the value of \(x\)

This same principle also applies to equations in which the base is \(0\) and \(-1\).

Given all of this, we need to consider a few different cases:

case i: \(x\) does not equal \(0\), \(1\) or \(-1\)
This means we can safely apply the above property to conclude that: \(3x^2-5x-7=2x^2-3x+1\)
Subtract \(2x^2\) from both sides of the equation to get: \(x^2-5x-7=-3x+1\)
Add \(3x\) to both sides of the equation to get: \(x^2-2x-7=1\)
Subtract \(1\) from both sides of the equation to get: \(x^2-2x-8=0\)
Factor: \((x-4)(x+2)=0\)
Solve: \(x=4\) and \(x=-2\)

case ii: \(x = 0\)
Since the base is one of the forbidden values (\(0\)), we can't apply the useful property above.
Instead, we can plug \(x=0\) into the original equation to determine whether it's a solution.
When we do this we get: \(0^{3(0^2)-5(0)-7}=0^{2(0^2)-3(0)+1}\)
When we simplify, we get: \(0^{-7}=0^{1}\)
Evaluate to get: \(\frac{1}{0}=0\).
Since \(\frac{1}{0}\) is undefined, \(x = 0\) is not a solution to the given equation.

Aside: Since the question asks us to find the SUM of all possible solutions, we didn't really need to check whether \(x=0\) is a solution, since it won't have any effect on the SUM. I just wanted to demonstrate the steps necessary to determining ALL possible solutions

case iii: \(x = 1\)
Once again, to determine whether \(x=1\) is a possible solution, we'll plug \(x=1\) into the original equation.
We get: \(1^{3(1^2)-5(1)-7}=1^{2(1^2)-3(1)+1}\)
Simplify: \(1^{-9}=1^{0}\)
Evaluate to get: \(1=1\). WORKS!
So, \(x=1\) is another solution

Aside: We don’t need to actually test \(x = 1\), since 1 raised to any power will always be 1. I’m just showing all possible steps.

case iv: \(x = -1\)
Once again, to determine whether \(x=-1\) is a possible solution, we'll plug \(x=-1\) into the original equation.
We get: \((-1)^{3(-1)^2)-5(-1)-7}=(-1)^{2(-1)^2)-3(-1)+1}\)
Simplify: \((-1)^{1}=(-1)^{6}\)
Evaluate to get: \(-1=1\). DOESN'T WORK
So, \(x=-1\) is NOT a solution

The sum of all possible solutions = \(4 + (-2) + 1 = 3\)

Answer: C

Cheers,
Brent