Solve each of the following quadratic equations:

Solution:

$\frac{4}{x}-3=\frac{5}{2 x+3}, \quad x \neq 0,-\frac{3}{2}$

$\Rightarrow \frac{4}{x}-\frac{5}{2 x+3}=3$

$\Rightarrow \frac{8 x+12-5 x}{x(2 x+3)}=3$

$\Rightarrow \frac{3 x+12}{2 x^{2}+3 x}=3$

$\Rightarrow \frac{x+4}{2 x^{2}+3 x}=1$

$\Rightarrow 2 x^{2}+3 x=x+4 \quad$ (Cross multiplication)

$\Rightarrow 2 x^{2}+2 x-4=0$

$\Rightarrow x^{2}+x-2=0$

$\Rightarrow x^{2}+2 x-x-2=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow(x+2)(x-1)=0$

$\Rightarrow x+2=0$ or $x-1=0$

$\Rightarrow x=-2$ or $x=1$

Hence, −2 and 1 are the roots of the given equation.