The repeated root of the equation $4x^3-12x^2-15x-4=0$  is

$f^1\left(x\right)=12x^2-24x-15$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{24\pm \sqrt{{\left(-24\right)}^2-4\left(-15\right)\left(12\right)}}{2\left(12\right)}$

$=\frac{24\pm \sqrt{576+720}}{24}$

$=\frac{24\pm \sqrt{1296}}{24}$

$=\frac{24+36}{24}\left(\text{or}\right)\frac{24-36}{24}$

$=\frac{60}{24}\left(\text{or}\right)\frac{-12}{24}$

$=\frac{15}{6}\left(\text{or}\right)\frac{-1}{2}$

Substitute $\frac{-1}{2}$  in the $f\left(x\right)=4x^3-12x^2-15x-4$

$=4{\left(-\frac{1}{2}\right)}^3-12{\left(-\frac{1}{2}\right)}^2-15\left(\frac{-1}{2}\right)-4$

$=4\left(-\frac{1}{8}\right)-12\left(\frac{1}{4}\right)+\frac{15}{2}-4$

$=-\frac{4}{8}-\left(\frac{6}{2}\right)+\frac{15}{2}-4$ $=4-4=0$

So $-\frac{1}{2}$  is a repeated root