Step 1 :The given equation is a radical equation, which means it contains square roots. To solve this type of equation, we usually isolate one of the square roots and then square both sides of the equation to eliminate the square root. However, in this case, we have negative signs in front of both square roots, which means the values under the square roots must be non-positive (i.e., negative or zero) because the square root of a positive number is always positive. Therefore, we need to find the values of x that make both \(-2x + 5\) and \(-3x - 1\) non-positive.
Step 2 :The solutions to the equations \(-2x + 5 \leq 0\) and \(-3x - 1 \leq 0\) are \(x = 2.5\) and \(x = -1/3\), respectively. This means that for the values of x in the interval \([-1/3, 2.5]\), both \(-2x + 5\) and \(-3x - 1\) are non-positive. Now, we need to substitute these values into the original equation and see if they satisfy it.
Step 3 :The solutions to the original equation when \(x = 2.5\) and \(x = -1/3\) are \(2 - 2.91547594742265i\) and \(4.38047614284762\), respectively. The first solution is a complex number, which means \(x = 2.5\) is not a solution to the original equation because the square root of a real number cannot be a complex number. The second solution is a real number, but it is not zero, which means \(x = -1/3\) is not a solution to the original equation either.
Step 4 :\(\boxed{\text{Therefore, the original equation has no solution.}}\)