What would be the temperature of the resultant mixture? The question in my textbook reads 100 g of

To freeze into ice, water needs a centre of condensation, around which the crystallization can take place. If the centre is absent or the water that is cooled is very pure then the water can be supercooled.
What's wrong here is that the water is not is its solid state, although it is supercooled. So as soon as the water gets the centre of condensation, the water will release energy and convert into ice and this energy released will be absorbed by the supercooled water, which will raise it's temperature.
$mL=Mc[0-<!--\; -\; -->(-<!--\; -\; -->10)]$
where M is the mass of supercooled water, m is the mass of ice formed, c is the specific heat of water.
$mL=10Mc$
The equilibrium will be achieved at $0^{\circ <!--\; \circ \; -->}$ C because once the freezing starts all the water will try to freeze because it's already below its freezing point. Until now, it was unable to freeze because it lacked a centre of condensation. Therefore, once that is achieved water begins to freeze instantaneously.
For instance, $100gms$ of water will require $1kcal$ of energy to reach $0^{\circ <!--\; \circ \; -->}C$ from $-<!--\; -\; -->{10}^{\circ <!--\; \circ \; -->}C$ and the water at $0^{\circ <!--\; \circ \; -->}C$ will release $8kcal$ of energy when all of the water is converted into ice.
Even if all the water freezes, the excess energy which was released when water freezes will melt the ice and the equilibrium will be achieved at $0^{\circ <!--\; \circ \; -->}C$ eventually. But this doesn't happen in this way in real world.
When all the components come at a common temperature then the equilibrium is achieved. In order for that to happen, the net heat lost and the net heat gained must be the same in an isolated system such that all the components achieve a common temperature.