Answer
$y=(\frac{-1+\sqrt 29}{2}, \frac{-1-\sqrt 29}{2})$
Work Step by Step
Step-1 : Add $7$ on both sides of the equation
$y^2+y=7$
Step -2 : Add the square of half of the co-efficient of $y$ to both sides.
Co-efficient of $y =1$
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 $y^2+y+\frac{1}{4}=7+\frac{1}{4}$
Step-3 Factor the trinomial and simplify the right hand side.
$(y+\frac{1}{2})^2=\frac{28+1}{4}$
$(y+\frac{1}{2})^2=\frac{29}{4}$
Step-4 Use the square root property and solve for $y$
$(y+\frac{1}{2})=±\sqrt\frac{29}{4}$
Step-5 Subtract $\frac{1}{2}$ on both the sides
$y=-\frac{1}{2}±\sqrt \frac{29}{4}$
Step-6 Simplify the right hand side
$y=-\frac{1}{2}±\frac{\sqrt 29}{2}$
$y=-\frac{1±\sqrt 29}{2}$
Therefore the solution set is $(\frac{-1+\sqrt 29}{2}, \frac{-1-\sqrt 29}{2})$
