Since one of the solutions is \(x=4\)

\(4^2-10(4)+13=k\)

\(16-40+13=k\)

\(k=-11\)

\(x^2-10x+13=-11\)

\(x^2-10x+24=0\)

Since \(x=4\) is one solution, one of the factors will be \((x-4)\)

\((x-4)(x-?)=0\)

We need a number in the place of the question mark which when added to \(-4\) gives us \(-10\) (the coefficient of the linear term) and which when multiplied to \(-4\) gives us \(24\) (the constant term). That number is \(-6\). Hence we get,

\((x-4)(x-6)=0\)

The other solution for \(x\) is \(6\).

\(x=6\)