Question 686982
To solve this, let's first set this problem up as an equation:

An integer plus three times the square of the integer is 10, can be set up as:

{{{x + 3x^2 = 10}}}

As you can see, this is a quadratic equation.  If we subtract the 10 from both sides, it will be set up in standard quadratic form, and we will be able to factor:

{{{3x^2 + x - 10 = 0}}}

{{{(3x - 5)(x + 2) = 0}}}

When we solve for x in the first set of parenthesis, we see that x equals 5/3.  5/3 is not an integer, therefore it cannot be part of the solution.

When we solve for x in the second set of parenthesis, we see that x = -2.  -2 IS an integer.

Therefore, -2 is our answer.  To validate, substitute -2 for x in our original equation:

{{{-2 + 3(-2^2) = 10}}} =

{{{-2 + 3(4) = 10}}} = 

{{{10 = 10}}}