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