The equation:
[tex]y^2+14y+49=0[/tex]
has the following values a=1, b=14 and c=49, hence the discriminant is:
[tex]14^2-4(1)(49)=196-196=0[/tex]
Since the discriminant is zero we have one solution of multiplicity 2 (this means we have one repeated solution)
Solving the quadratic equation with the general formula:
[tex]\begin{gathered} y=\frac{-14\pm\sqrt[]{14^2-4(1)(49)}}{2(1)} \\ y=\frac{-14\pm\sqrt[]{0}}{2} \\ y=\frac{-14}{2} \\ y=-7 \end{gathered}[/tex]
Therefore the solution is y=-7
