Brinell Hardness Formula

Note: Wolfram/Alpha gives these 2 results: I have changed "BHN" on the LHS to just "B"

d = -(2 sqrt(pi B D^2 P-P^2))/(pi B D) and,

d = (2 sqrt(pi B D^2 P-P^2))/(pi B D)

How do I rearrange the formula BHN = (2*P) / [ (pi*D)(D - sqrt(D2-d2)) ] to make 'd' the subject?
Thank you

\(\begin{array}{rcll} \text{BHN} &=& \frac{2\cdot P} { \pi \cdot D \cdot (D - \sqrt{D^2-d^2}) } \qquad & | \qquad \cdot (D - \sqrt{D^2-d^2})\\ \text{BHN}\cdot (D - \sqrt{D^2-d^2}) &=& \frac{2\cdot P} { \pi \cdot D } \qquad & | \qquad : \text{BHN}\\ D - \sqrt{D^2-d^2} &=& \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \qquad & | \qquad \cdot(-1)\\ -D + \sqrt{D^2-d^2} &=& -\frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \qquad & | \qquad +D\\ \sqrt{D^2-d^2} &=& D-\frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \qquad & | \qquad \text{square both sides}\\ D^2-d^2 &=& \left( D-\frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right)^2\\ D^2-d^2 &=& D^2 -2\cdot D \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right) + \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right)^2\\ -d^2 &=& -2\cdot D \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right) + \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right)^2 \qquad & | \qquad \cdot(-1)\\ d^2 &=& 2\cdot D \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right) - \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right)^2\\ d^2 &=& \frac{4\cdot D \cdot P} { \pi \cdot D \cdot \text{BHN} } - \frac{4\cdot P^2} { \left( \pi \cdot D \cdot \text{BHN}\right)^2 } \\ d^2 &=& \frac{4\cdot D \cdot P} { \pi \cdot D \cdot \text{BHN} } \cdot \frac{\pi \cdot D \cdot \text{BHN}}{\pi \cdot D \cdot \text{BHN}} - \frac{4\cdot P^2} { \left( \pi \cdot D \cdot \text{BHN}\right)^2 } \\ d^2 &=& \frac{4\cdot D \cdot P\cdot \pi \cdot D \cdot \text{BHN}} { \left( \pi \cdot D \cdot \text{BHN}\right)^2 } - \frac{4\cdot P^2} { \left( \pi \cdot D \cdot \text{BHN}\right)^2 } \\ d^2 &=& \frac{4}{\left( \pi \cdot D \cdot \text{BHN}\right)^2} \cdot \left( D^2 \cdot P\cdot \pi \cdot \text{BHN} - P^2 \right) \qquad & | \qquad \pm\sqrt{}\\ d &=& \pm \frac{2}{ \pi \cdot D \cdot \text{BHN} } \cdot \sqrt{ D^2 \cdot P\cdot \pi \cdot \text{BHN} - P^2 } \\ d &=& \pm \frac{2\cdot \sqrt{ D^2 \cdot P\cdot \pi \cdot \text{BHN} - P^2 } }{ \pi \cdot D \cdot \text{BHN} } \\\\ \mathbf{d_1 } &\mathbf{=}& \mathbf{ + \frac{2\cdot \sqrt{ D^2 \cdot P\cdot \pi \cdot \text{BHN} - P^2 } }{ \pi \cdot D \cdot \text{BHN} } }\\\\ \mathbf{d_2 } &\mathbf{=}& \mathbf{ - \frac{2\cdot \sqrt{ D^2 \cdot P\cdot \pi \cdot \text{BHN} - P^2 } }{ \pi \cdot D \cdot \text{BHN} } } \end{array}\)