Explanation:
since it is a quadratic equation, we know it must have 2 solutions.
132 is a bit larger than 10².
so, let's try x = 10 :
10 + 10² = 132
10 + 100 = 132
110 = 132
not correct, but close.
the real solution must be a bit larger than x = 10.
we know, that 12² = 144. that is already too large (as the sum with x must be only 132).
so, let's try x = 11
11 + 11² = 132
11 + 121 = 132
132 = 132
correct !
so, x = 11 is one solution.
for the second solution, it is very often that it has the opposite sign to the first solution, but its absolute value is relatively close to the first solution.
so, when thinking negative values, x² has to go higher than 132, so that x (with its negative value) can bring it back down to 132.
what we just thought about +12 might have some merit for -12.
let's try x = -12 :
-12 + (-12)² = 132
-12 + 144 = 132
132 = 132
correct !
so, x = -12 is the second solution.
summary :
x = 11
x = -12
are the 2 solutions.