The correct option is option (c) $\frac{3}{\sqrt{2}}, \frac{\sqrt{2}}{4}$
$f(x)=4 x^{2}+5 \sqrt{2} x-3=0$
$\Rightarrow 4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3=0$
$\Rightarrow 2 \sqrt{2} \mathrm{x}(\sqrt{2} \mathrm{x}+3)-1(\sqrt{2} \mathrm{x}+3)=0$
$\Rightarrow(\sqrt{2} x+3)(2 \sqrt{2} x-1)=0$
$\Rightarrow \mathrm{x}=-\frac{3}{\sqrt{2}}$ or $\mathrm{x}=\frac{1}{2 \sqrt{2}}$
$\Rightarrow \mathrm{x}=-\frac{3}{\sqrt{2}}$ or $\mathrm{x}=\frac{1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{4}$