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LHS;
Consider the quadratic equation;

$a d^{2}\left(\frac{a x}{b}+\frac{2 c}{d}\right) x+b c^{2}=0$

Put $x=-\frac{b c}{a d}$ in the given equation.

$a d^{2}\left[\frac{a\left(-\frac{b c}{a d}\right)}{b}+\frac{2 c}{d}\right]\left(-\frac{b c}{a d}\right)+b c^{2}$

$=\frac{a d^{2}}{b d}\left[a d\left(-\frac{b c}{a d}\right)+2 b c\right]\left(-\frac{b c}{a d}\right)+b c^{2}$

$=\frac{a d^{2}}{b d}[-b c+2 b c]\left(-\frac{b c}{a d}\right)+b c^{2}$

$=\frac{a d^{2}}{b d}(b c)\left(-\frac{b c}{a d}\right)+b c^{2}$

$=-\frac{a b^{2} c^{2} d^{2}}{a b d^{2}}+b c^{2}$

$=-b c^{2}+b c^{2}$

$=0=\mathrm{RHS}$

Hence, $x=-\frac{b c}{a d}$ is a solution to the given quadratic equation.