Rearranging the formula?

p = (n² + a) / (n + a)

Start with multiplying both sides by the denominator:

p(n + a) = n² + a

Now distribute the p through the parenthesis on the left side:

np + ap = n² + a

Now set it up like a quadratic, moving everything over to one side:

0 = n² - np + a - ap

Now put this into quadratic equation where:

a = 1, b = -p, c = (a - ap)

n = [-b ± √(b² - 4ac)] / (2a)

n = [-(-p) ± √((-p)² - 4(1)(a - ap))] / (2(1))

n = [p ± √(p² - 4(a - ap))] / 2

n = [p ± √(p² - 4a + 4ap)] / 2

And that is as simplified as you can get.

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As a test, let's plug in some numbers into p and a, then solve for n from the starting equation and solve for n using my derived equation, and we should get the same answer:

Let's say: p = 0, a = 5...

p = (n² + a) / (n + a)

0 = (n² + 5) / (n + 5)

0 = n² + 5

n² = -5

n = ±√(5)i

Now if we put the same values into the derived equation:

n = [p ± √(p² - 4a + 4ap)] / 2

n = [0 ± √(0² - 4(5) + 4(5)(0))] / 2

n = ±√(-20) / 2

n = ±√(-4 * 5) / 2

n = ±2i√(5) / 2

n = ±i√5

we get the same answer.

pn-pa=n^2+a

n^2-pn+pa+a=0

n^2-pn+a(p+1)=0 use the quadratic formula for simplicity...

n=[p±(p^2-4a(p+1))^(1/2)]/2