..pls solve this q6 and 7 by *completing the square method*

Dear Student,

6.
$$\begin{gathered} 3x^2-16+5=0 \\ {\left(\sqrt{3}x\right)}^2-2\left(\sqrt{3}x\right)\left(\frac{8}{\sqrt{3}}\right)+\frac{16}{3}-\frac{16}{3}+5=0 \\ {\left(\sqrt{3}x-\frac{8}{\sqrt{3}}\right)}^2-\frac{1}{3}=0 \\ {\left(\sqrt{3}x-\frac{8}{\sqrt{3}}\right)}^2-{\left(\frac{1}{\sqrt{3}}\right)}^2=0 \\ Since, a^2-b^2=(a-b)(a+b) \\ \left(\sqrt{3}x-\frac{8}{\sqrt{3}}-\frac{1}{\sqrt{3}}\right)\left(\sqrt{3}x-\frac{8}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right)=0 \\ \left(\frac{3x-9}{\sqrt{3}}\right)\left(\frac{3x-7}{\sqrt{3}}\right)=0 \\ x=3 or x=\frac{7}{3} \end{gathered}$$

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