$\begin{array}{l}
7 x^{2}+3 x-4=0 \\
\Rightarrow 49 x^{2}+21 x-28=0
\end{array}$
$\Rightarrow 49 x^{2}+21 x=28$
$\Rightarrow(7 x)^{2}+2 \times 7 x \times \frac{3}{2}+\left(\frac{3}{2}\right)^{2}=28+\left(\frac{3}{2}\right)^{2} \quad$ [Adding $\left(\frac{3}{2}\right)^{2}$ on both side]
$\Rightarrow\left(7 x+\frac{3}{2}\right)^{2}=28+\frac{9}{4}=\frac{121}{4}=\left(\frac{11}{2}\right)^{2}$
$\Rightarrow 7 x+\frac{3}{2}=\pm \frac{11}{2} \quad$ (Taking square root on both sides)
$\Rightarrow 7 x+\frac{3}{2}=\frac{11}{2}$ or $7 x+\frac{3}{2}=-\frac{11}{2}$
$\Rightarrow 7 x=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4$ or $7 x=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7$
$\Rightarrow x=\frac{4}{7}$ or $x=-1$

Hence, $\frac{4}{7}$ and $-1$ are the roots of the given equation.