Answer:
The solution is:
[tex]x=\frac{\sqrt{29}+7}{2},\:x=\frac{-\sqrt{29}+7}{2}[/tex]
Step-by-step explanation:
Given the equation
[tex]x^2\:-7x=\:-5[/tex]
[tex]\mathrm{Add\:}a^2=\left(-\frac{7}{2}\right)^2\mathrm{\:to\:both\:sides}[/tex]
[tex]x^2-7x+\left(-\frac{7}{2}\right)^2=-5+\left(-\frac{7}{2}\right)^2[/tex]
[tex]x^2-7x+\left(-\frac{7}{2}\right)^2=\frac{29}{4}[/tex]
Apply perfect square rule
[tex]\left(x-\frac{7}{2}\right)^2=\frac{29}{4}[/tex]
[tex]\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}[/tex]
solving
[tex]x-\frac{7}{2}=\sqrt{\frac{29}{4}}[/tex]
[tex]x-\frac{7}{2}=\frac{\sqrt{29}}{\sqrt{4}}[/tex]
[tex]x-\frac{7}{2}=\frac{\sqrt{29}}{2}[/tex]
[tex]\mathrm{Add\:}\frac{7}{2}\mathrm{\:to\:both\:sides}[/tex]
[tex]x-\frac{7}{2}+\frac{7}{2}=\frac{\sqrt{29}}{2}+\frac{7}{2}[/tex]
[tex]x=\frac{\sqrt{29}+7}{2}[/tex]
also solving
[tex]x-\frac{7}{2}=-\sqrt{\frac{29}{4}}[/tex]
[tex]\mathrm{Add\:}\frac{7}{2}\mathrm{\:to\:both\:sides}[/tex]
[tex]x-\frac{7}{2}+\frac{7}{2}=-\frac{\sqrt{29}}{2}+\frac{7}{2}[/tex]
[tex]x=\frac{-\sqrt{29}+7}{2}[/tex]
Thus, the solution is:
[tex]x=\frac{\sqrt{29}+7}{2},\:x=\frac{-\sqrt{29}+7}{2}[/tex]
