3}}{5} + 0 = 0\)Due to the high input impedance of the Op-Amp, the input CURRENT to the Op-amp will be 0.\(V = \frac{{{V_{out}} + 6}}{3}\)For an upper saturated VOLTAGE output of VOUT = +15 V, we get the THRESHOLD voltage V of:\(V =V_{in}= \frac{{15 + 6}}{3} \)\(V= \frac{{21}}{3}=7~V\)For a lower saturated voltage of Vout = -15 V, we get the threshold voltage V of:\(V = V_{in}=\frac{{ - 15 + 6}}{3}\)\(V== \frac{{ - 9}}{3}=-3~V\)