$2 x^{2}+5 x-3=0$
$\Rightarrow 4 x^{2}+10 x-6=0 \quad$ (Multiplying both sides by 2) $\Rightarrow 4 x^{2}+10 x=6$
$\Rightarrow(2 x)^{2}+2 \times 2 x \times \frac{5}{2}+\left(\frac{5}{2}\right)^{2}=6+\left(\frac{5}{2}\right)^{2} \quad$ [Adding $\left(\frac{5}{2}\right)^{2}$ on both sides $]$
$\Rightarrow\left(2 x+\frac{5}{2}\right)^{2}=6+\frac{25}{4}=\frac{24+25}{4}=\frac{49}{4}=\left(\frac{7}{2}\right)^{2}$
$\Rightarrow 2 x+\frac{5}{2}=\pm \frac{7}{2} \quad$ (Taking square root on both sides)
$\Rightarrow 2 x+\frac{5}{2}=\frac{7}{2}$ or $2 x+\frac{5}{2}=-\frac{7}{2}$
$\Rightarrow 2 x=\frac{7}{2}-\frac{5}{2}=\frac{2}{3}=1$ or $2 x=-\frac{7}{2}-\frac{5}{2}=-\frac{12}{2}=-6$
$x=\frac{1}{2}$ or $x=-3$

Hence, $\frac{1}{2}$ and $-3$ are the roots of the given equation.