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 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}

The equation can now be written as

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

and has, after taking the square root, the solutions:

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

• \displaystyle y+1 = -\sqrt{16} = -4\,\textrm{,}\ which gives \displaystyle y=-1-4=-5\,\textrm{.}

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

• y = -5: \displaystyle \ \text{LHS} = (-5)^{2} + 2\cdot (-5)-15 = 25-10-15 = 0 = \text{RHS,}

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