kindly help me to solve this equation
how he calculated vlue of d to be 16.1
$\displaystyle RHS=0.022\times10^{-6}\times 3\times10^8=6.6$
Therefore, you have:
$\displaystyle 2\sqrt{\bigg(\dfrac{d}{2}\bigg)^2+8^2}-d= 6.6$
$\displaystyle 2\sqrt{\bigg(\dfrac{d^2}{4}\bigg)+64}=6.6+d$
squaring both sides gives you:
$\displaystyle \bigg[2\sqrt{\bigg(\dfrac{d^2}{4}\bigg)+64}\bigg]^2=(6.6+d)^2$
$\displaystyle 4 \cdot \bigg(\dfrac{d^2}{4}+64\bigg) = 6.6^2+2d(6.6)+ d^2$
$\displaystyle d^2+256=43.56+13.2d+d^2$
the $\displaystyle d^2$ terms cancel each other out and simplification gives you the value of d.