Solve the radical equation: 3sqrt(4x−3) − sqrt(5x+1) = 5

We start with the given equation:

$$3\sqrt{4x-3} - \sqrt{5x+1} = 5$$

Add $\sqrt{5x+1}$ to both sides to isolate the first radical term:

$$3\sqrt{4x-3} = 5 + \sqrt{5x+1}$$

Squaring both sides eliminates the outermost radical on the left and expands the right side using $(a+b)^2 = a^2 + 2ab + b^2$:

$$(3\sqrt{4x-3})^2 = (5 + \sqrt{5x+1})^2$$ $$9(4x-3) = 5^2 + 2(5)\sqrt{5x+1} + (\sqrt{5x+1})^2$$ $$36x - 27 = 25 + 10\sqrt{5x+1} + 5x + 1$$

Combine like terms on the right side:

$$36x - 27 = 26 + 5x + 10\sqrt{5x+1}$$

Move all non-radical terms to the left side:

$$36x - 5x - 27 - 26 = 10\sqrt{5x+1}$$ $$31x - 53 = 10\sqrt{5x+1}$$

For this equation to have real solutions, we must have $31x - 53 \ge 0$, which means $x \ge \frac{53}{31} \approx 1.71$.

$$(31x - 53)^2 = (10\sqrt{5x+1})^2$$

Expand the left side using $(a-b)^2 = a^2 - 2ab + b^2$:

$$(31x)^2 - 2(31x)(53) + 53^2 = 100(5x+1)$$ $$961x^2 - 3286x + 2809 = 500x + 100$$

$$961x^2 - 3286x - 500x + 2809 - 100 = 0$$ $$961x^2 - 3786x + 2709 = 0$$

We use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a = 961$, $b = -3786$, $c = 2709$.
Calculate the discriminant $\Delta = b^2 - 4ac$:

$$\Delta = (-3786)^2 - 4(961)(2709)$$ $$\Delta = 14333796 - 10413396$$ $$\Delta = 3920400$$

Now find $\sqrt{\Delta}$:

$$\sqrt{3920400} = 1980$$

Substitute these values into the quadratic formula:

$$x = \frac{-(-3786) \pm 1980}{2(961)}$$ $$x = \frac{3786 \pm 1980}{1922}$$

This gives two potential solutions:

$$x_1 = \frac{3786 + 1980}{1922} = \frac{5766}{1922} = 3$$ $$x_2 = \frac{3786 - 1980}{1922} = \frac{1806}{1922} = \frac{903}{961}$$

The problem states that $x$ must be an integer. From our solutions, $x_1 = 3$ is an integer, while $x_2 = \frac{903}{961}$ is not.
Also, recall the condition $x \ge \frac{53}{31} \approx 1.71$.
$x_1 = 3$ satisfies this condition.
$x_2 = \frac{903}{961} \approx 0.94$, which does not satisfy $x \ge 1.71$, so it is an extraneous solution.

Let's verify $x=3$ in the original equation:

$$3\sqrt{4(3)-3} - \sqrt{5(3)+1} = 5$$ $$3\sqrt{12-3} - \sqrt{15+1} = 5$$ $$3\sqrt{9} - \sqrt{16} = 5$$ $$3(3) - 4 = 5$$ $$9 - 4 = 5$$ $$5 = 5$$

The solution $x=3$ is correct.

The final answer is $\mathbf{3}$.