From Förberedande kurs i matematik 1

The first step when we solve the second-degree equation is to complete the square on the left-hand side:

\displaystyle y^{2}+2y-15=\left( y+1 \right)^{2}-1^{2}-15=\left( y+1 \right)^{2}-16.

The equation can now be written as

\displaystyle \left( y+1 \right)^{2}=16

and has, after taking the square root, the solutions

\displaystyle y+1=\sqrt{16}=4 which gives \displaystyle y=-1+4=3

\displaystyle y+1=-\sqrt{16}=-4 which gives \displaystyle y=-1-4=-5

A quick check shows that \displaystyle y=-\text{5 } and \displaystyle y=\text{3 } satisfy the equation:

\displaystyle y=-\text{5 }: LHS= \displaystyle \left( -5 \right)^{2}+2\centerdot \left( -5 \right)-15=25-10-15=0 = RHS

\displaystyle y=\text{3 }: LHS= \displaystyle 3^{2}+2\centerdot 3-15=9+6-15=0 = RHS