What is the answer to solvefor n,y=in(x^2+2x+5) ?

The answer to solvefor n,y=in(x^2+2x+5) is n=-(iy)/(x^2+2x+5);\quad x\ne-1+2i,x\ne-1-2i

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solvefor n,y=in(x^2+2x+5)

solve for

n, y = in (x^{2} + 2 x + 5)

n = - \frac{i y}{x ^{2} + 2 x + 5}; x \neq = - 1 + 2 i, x \neq = - 1 - 2 i

Solution steps

y = in (x^{2} + 2 x + 5)

Switch sides

in (x^{2} + 2 x + 5) = y

Divide both sides by

i (x^{2} + 2 x + 5); x \neq = - 1 + 2 i, x \neq = - 1 - 2 i

n = - \frac{i y}{x ^{2} + 2 x + 5}; x \neq = - 1 + 2 i, x \neq = - 1 - 2 i

\frac{x}{4} + 16 = 2 x

\frac{3}{5} x + 2 = \frac{2}{3} x - 4

2 y - 4 = 3 y - 5

\frac{4 x - 5}{3} = 2 x + 9

\frac{x - 1}{3} + \frac{2 x + 1}{4} = 3

What is the answer to solvefor n,y=in(x^2+2x+5) ?
The answer to solvefor n,y=in(x^2+2x+5) is n=-(iy)/(x^2+2x+5);\quad x\ne-1+2i,x\ne-1-2i