Answer
$x=(\frac{7+\sqrt 53}{2}, \frac{7-\sqrt 53}{2})$
Work Step by Step
Step-1 : Add $1$ on both sides of the equation
$x^2-7x=1$
Step -2 : Add the square of half of the co-efficient of $x$ to both sides.
Co-efficient of $x =-7$
Half of $-7 = \frac{1}{2}×-7=\frac{-7}{2}$
Square of $\frac{-7}{2}$ is $\frac{-7}{2}×\frac{-7}{2}=\frac{49}{4}$
The equation becomes $x^2-7x+\frac{49}{4}=1+\frac{49}{4}$
Step-3 Factor the trinomial and simplify the right hand side.
$(x-\frac{7}{2})^2=\frac{4+49}{4}$
$(x-\frac{7}{2})^2=\frac{53}{4}$
Step-4 Use the square root property and solve for $x$
$(x-\frac{7}{2})=±\sqrt\frac{53}{4} $
Step-5 Add $\frac{7}{2}$ on both the sides
$x=\frac{7}{2}±\sqrt \frac{53}{4}$
Step-6 Simplify the right hand side
$x=\frac{7}{2}±\frac{\sqrt 53}{2}$
$x=\frac{7±\sqrt 53}{2}$
Therefore the solution set is $(\frac{7+\sqrt 53}{2}, \frac{7-\sqrt 53}{2})$
