S = \(\sqrt{t^2 - 4t + 4}\)

S= t- 4t + 4

t- 4t + 4 - S= 0

Using \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substituting, we have;

Using \(t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4 - S^2)}}{2(1)}\)

\(t = \frac{4 \pm \sqrt{16 - 4(4 - S^2)}}{2}\)

\(t = \frac{4 \pm \sqrt{16 - 16 + 4S^2}}{2}\)

\(t = \frac{4 \pm \sqrt{4S^2}}{2}\)

\(t = \frac{2(2 \pm S)}{2}\)

Hence t = 2 + S or t = 2 - S