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$\\Solution:\\ Given\ complex\ number= 5-12i\\\\ let\ x+yi=\sqrt{ 5-12i}\\ \\Squaring\ on\ both\ sides\\\\ x^2+(yi)^2+2xyi= 5-12i\\\\ (x^2-y^2)+2xyi= 5-12i\\\\ On\ comparing\ the\ corresponding\ terms\\ x^2-y^2=5\rightarrow (1)\ \ \ \ \ \ \ 2xy=-12\ \ \Rightarrow xy=-6\rightarrow (2)\\\\ Squaring\ equation\ (1)\\\\ \Rightarrow (x^2-y^2)^2=25\\\\ \Rightarrow (x^2+y^2)^2-4x^2y^2=25\ \ \ [\because (a-b)^2=(a+b)^2-4ab]\\\\ \Rightarrow (x^2+y^2)^2=25+4\times36\\\\ \Rightarrow (x^2+y^2)^2=169\\\\ \Rightarrow x^2+y^2=13\rightarrow (3)\\\\ Adding (1)\ and\ (2)\ we\ will\ get\\\\ \Rightarrow 2x^2=18\\\\ \Rightarrow x^2=9\\\\ \Rightarrow x=\pm 3\ \ and\\ \Rightarrow y=\pm2\ \ \ [\because From (2)]\\\\ Since\ imaginary\ part\ is\ negative then\\ the\ square\ root \ of\ 5-12i\ is\pm(3-2i)$

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