Read the block diagram. It is a unity negative-feedback system with two cascaded blocks in the forward path:

\(G_1(s)=\dfrac{3}{s+15}, \qquad G_2(s)=\dfrac{15}{s+1}\)

Step 1 — combine the cascade (series blocks multiply):

\(G(s)=\dfrac{3}{s+15}\times\dfrac{15}{s+1}=\dfrac{45}{(s+15)(s+1)}\)

Step 2 — write the characteristic equation. For unity feedback the closed-loop transfer function is \(\dfrac{G}{1+G}\), so the characteristic equation is the denominator set to zero:

\(1+G(s)=0 \Rightarrow 1+\dfrac{45}{(s+15)(s+1)}=0\)

Step 3 — clear the fraction and expand.

\((s+15)(s+1)+45=0\)

\(s^{2}+16s+15+45=0\)

\(s^{2}+16s+60=0\)

Step 4 — solve the quadratic.

\(s=\dfrac{-16\pm\sqrt{16^{2}-4(60)}}{2}=\dfrac{-16\pm\sqrt{256-240}}{2}=\dfrac{-16\pm 4}{2}\)

\(s=-6 \quad\text{and}\quad s=-10\)

(Quick check by factorising: \(s^{2}+16s+60=(s+6)(s+10)\) ✓, since 6 + 10 = 16 and 6 × 10 = 60.)

These match statements B (root = −6) and D (root = −10).

Why the positive roots in A and C are impossible here. All the coefficients of \(s^{2}+16s+60\) are positive, so by Descartes' rule (and by Routh–Hurwitz) there can be no root with a positive real part. Physically, the closed-loop poles at −6 and −10 are both in the left half plane, so this system is stable and its response is over-damped (two distinct real poles), with the slower pole at −6 dominating the settling time.

Note the shift caused by feedback. The open-loop poles were at −1 and −15; closing the loop with gain 45 pulls them together to −6 and −10 — a nice illustration of how root-locus branches move as loop gain rises.

Hence, the roots of the characteristic equation are −6 and −10, i.e. B and D only.