If c = 2*10^6 and y = 2*10^5

If c = 2*10^6 and y = 2*10^5 
w^2 = c*y/c-y 
Work out the value of w. Give your answer in standard form.

w^2 =[2*10^6] x [2*10^5] / [2*10^6  -  2*10^5]

w^2 =[4*10^11] / [1.8*10^6]

w^2 =222,222.22......

w = sqrt(222,222.22...)

w ~ =+,- 1,000sqrt(2)/3 =+,- 471.40....

If     w  =  $c\cdot\frac{y}{c}-y$     , then this is how to do it.

c  =  2 · 10⁶

y  =  2 · 10⁵

$w^2\,=\,c\cdot\frac{y}{c}-y$                                      Plug in the given values for  c  and  y .

$w^2\,=\,2\cdot10^6\cdot\frac{2\cdot10^5}{2\cdot10^6}-2\cdot10^5 \\ w^2\,=\, 2\cdot10^6\cdot\frac{10^5}{10^6}-2\cdot10^5 \\ w^2  \,=\,   2\cdot10^6\cdot\frac{1}{10}-2\cdot10^5 \\ w^2  \,=\,   2\cdot10^5-2\cdot10^5 \\ w^2\,=\,0\\w\,\ \,=\,0$              Reduce the fraction by  2 .

Or....if     $w^2\,=\,\frac{c\cdot y}{c-y}$     , then this is how to do it.

$w^2\,=\,\frac{2\cdot10^6\cdot2\cdot10^5}{2\cdot10^6-2\cdot10^5} \,=\, \frac{4\cdot10^{11}}{20\cdot10^5-2\cdot10^5} \,=\,\frac{4\cdot10^{11}}{18\cdot10^5}\,=\,\frac{2\cdot10^6}{9}\\~\\ w\,=\,\pm\sqrt{\frac{2\cdot10^6}{9}}\,=\,\pm\frac{10^3\sqrt2}{3}\,=\,\pm\frac{1000\sqrt2}{3}$

The second one is the same answer as Guest's.