This is a quartic equation. It would be complex to expand the equation to its general form and solve in conventional methods.

Here let’s use substitution method to find the solutions.

Let

Then,

(4x+3)^2(2x+1)(x+1)

=t^2(t-1)(t+1)

=72

Expanding the equation in terms of t gives biquadratic equation,

t^4-t^2-72=0

Factorizing the LHS

Then we get 4 solutions for t

(t^2-9)(t^2+8)=0

\begin{cases} t_1=3 \\ t_2=-3 \\ t_3=2\sqrt{2}i \\ t_4=-2\sqrt{2}i \end{cases}

Substitute the values of t to (1), then we obtain 4 solutions of the original equation.

\begin{cases} x_1=0 \\ x_2=-\dfrac{3}{2} \\ x_3=-\dfrac{3}{4}+\dfrac{\sqrt{2}i }{2} \\ x_4=-\dfrac{3}{4}-\dfrac{\sqrt{2}i }{2} \end{cases}