Let

$x+iy=\sqrt{\left(-3+4i\right)\left(8+6i\right)}$, $x,y\in R$

$\Rightarrow {\left(x+iy\right)}^2=\left(-3+4i\right)\left(8+6i\right)$

$\Rightarrow \left(x^2-y^2\right)+i\left(2xy\right)=-48+14i$

On comparing rea and imaginary part, we get

$x^2-y^2=-48 ....\left(\mathrm{i}\right)$

$xy=7 ....\left(\mathrm{ii}\right)$

$\Rightarrow x^2{y}^2=49$

$\Rightarrow x^2\left(x^2+48\right)=49$

$\Rightarrow x^4+48x^2-49=0$

$\Rightarrow x^4+49x^2-x^2-49=0$

$\Rightarrow \left(x^2-1\right)\left(x^2+49\right)=0$

$\Rightarrow x=\pm 1, \pm 7i$

Since, $x, y\in R$, so

$x=\pm 1\Rightarrow y=\pm 7$

Hence,

$\sqrt{\left(-3+4i\right)\left(8+6i\right)}=\pm \left(1+7i\right)$