Remove parentheses.
\[=2{(\imath nfty)}^{2}+\imath nfty+5+\imath nfty-3\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[=2{\imath }^{2}{n}^{2}{f}^{2}{t}^{2}{y}^{2}+\imath nfty+5+\imath nfty-3\]
Use Square Rule: \({i}^{2}=-1\).
\[=2\times -1\times {n}^{2}{f}^{2}{t}^{2}{y}^{2}+\imath nfty+5+\imath nfty-3\]
Move all terms to one side.
\[2\times -1\times {n}^{2}{f}^{2}{t}^{2}{y}^{2}+\imath nfty+5+\imath nfty-3=0\]
Simplify \(2\times -1\times {n}^{2}{f}^{2}{t}^{2}{y}^{2}\) to \(-2{n}^{2}{f}^{2}{t}^{2}{y}^{2}\).
\[-2{n}^{2}{f}^{2}{t}^{2}{y}^{2}+\imath nfty+5+\imath nfty-3=0\]
Simplify \(-2{n}^{2}{f}^{2}{t}^{2}{y}^{2}+\imath nfty+5+\imath nfty-3\) to \(-2{n}^{2}{f}^{2}{t}^{2}{y}^{2}+6\imath nfty-3\).
\[-2{n}^{2}{f}^{2}{t}^{2}{y}^{2}+6\imath nfty-3=0\]
Use the Quadratic Formula.
In general, given \(a{x}^{2}+bx+c=0\), there exists two solutions where:
\[x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\]
In this case, \(a=-2{f}^{2}{t}^{2}{y}^{2}\), \(b=6\imath fty\) and \(c=-3\).
\[{n}^{}=\frac{-6\imath fty+\sqrt{{(6\imath fty)}^{2}-4\times -2{f}^{2}{t}^{2}{y}^{2}\times -3}}{2\times -2{f}^{2}{t}^{2}{y}^{2}},\frac{-6\imath fty-\sqrt{{(6\imath fty)}^{2}-4\times -2{f}^{2}{t}^{2}{y}^{2}\times -3}}{2\times -2{f}^{2}{t}^{2}{y}^{2}}\]
Simplify.
\[n=\frac{-6\imath fty+2\sqrt{15}\imath fty}{-4{f}^{2}{t}^{2}{y}^{2}},\frac{-6\imath fty-2\sqrt{15}\imath fty}{-4{f}^{2}{t}^{2}{y}^{2}}\]
\[n=\frac{-6\imath fty+2\sqrt{15}\imath fty}{-4{f}^{2}{t}^{2}{y}^{2}},\frac{-6\imath fty-2\sqrt{15}\imath fty}{-4{f}^{2}{t}^{2}{y}^{2}}\]
Simplify solutions.
\[n=\frac{(3-\sqrt{15})\imath }{2fty},\frac{(3+\sqrt{15})\imath }{2fty}\]
Break down the problem into these 2 equations.
\[n=\frac{(3-\sqrt{15})\imath }{2fty}\]
\[n=\frac{(3+\sqrt{15})\imath }{2fty}\]
Solve the 1st equation: \(n=\frac{(3-\sqrt{15})\imath }{2fty}\).
Multiply both sides by \(2fty\).
\[n\times 2fty=(3-\sqrt{15})\imath \]
Regroup terms.
\[2nfty=(3-\sqrt{15})\imath \]
Divide both sides by \(2\).
\[nfty=\frac{(3-\sqrt{15})\imath }{2}\]
Divide both sides by \(n\).
\[fty=\frac{\frac{(3-\sqrt{15})\imath }{2}}{n}\]
Simplify \(\frac{\frac{(3-\sqrt{15})\imath }{2}}{n}\) to \(\frac{(3-\sqrt{15})\imath }{2n}\).
\[fty=\frac{(3-\sqrt{15})\imath }{2n}\]
Divide both sides by \(f\).
\[ty=\frac{\frac{(3-\sqrt{15})\imath }{2n}}{f}\]
Simplify \(\frac{\frac{(3-\sqrt{15})\imath }{2n}}{f}\) to \(\frac{(3-\sqrt{15})\imath }{2nf}\).
\[ty=\frac{(3-\sqrt{15})\imath }{2nf}\]
Divide both sides by \(t\).
\[y=\frac{\frac{(3-\sqrt{15})\imath }{2nf}}{t}\]
Simplify \(\frac{\frac{(3-\sqrt{15})\imath }{2nf}}{t}\) to \(\frac{(3-\sqrt{15})\imath }{2nft}\).
\[y=\frac{(3-\sqrt{15})\imath }{2nft}\]
\[y=\frac{(3-\sqrt{15})\imath }{2nft}\]
Solve the 2nd equation: \(n=\frac{(3+\sqrt{15})\imath }{2fty}\).
Multiply both sides by \(2fty\).
\[n\times 2fty=(3+\sqrt{15})\imath \]
Regroup terms.
\[2nfty=(3+\sqrt{15})\imath \]
Divide both sides by \(2\).
\[nfty=\frac{(3+\sqrt{15})\imath }{2}\]
Divide both sides by \(n\).
\[fty=\frac{\frac{(3+\sqrt{15})\imath }{2}}{n}\]
Simplify \(\frac{\frac{(3+\sqrt{15})\imath }{2}}{n}\) to \(\frac{(3+\sqrt{15})\imath }{2n}\).
\[fty=\frac{(3+\sqrt{15})\imath }{2n}\]
Divide both sides by \(f\).
\[ty=\frac{\frac{(3+\sqrt{15})\imath }{2n}}{f}\]
Simplify \(\frac{\frac{(3+\sqrt{15})\imath }{2n}}{f}\) to \(\frac{(3+\sqrt{15})\imath }{2nf}\).
\[ty=\frac{(3+\sqrt{15})\imath }{2nf}\]
Divide both sides by \(t\).
\[y=\frac{\frac{(3+\sqrt{15})\imath }{2nf}}{t}\]
Simplify \(\frac{\frac{(3+\sqrt{15})\imath }{2nf}}{t}\) to \(\frac{(3+\sqrt{15})\imath }{2nft}\).
\[y=\frac{(3+\sqrt{15})\imath }{2nft}\]
\[y=\frac{(3+\sqrt{15})\imath }{2nft}\]
Collect all solutions.
\[y=\frac{(3-\sqrt{15})\imath }{2nft},\frac{(3+\sqrt{15})\imath }{2nft}\]
y=((3-sqrt(15))*IM)/(2*n*f*t),((3+sqrt(15))*IM)/(2*n*f*t)
