For what values of $j$ does the equation $(2x + 7)(x - 4) = -31 + jx$ have exactly one real solution? Express your answer as a list of numbers, separated by commas.
When the equation is expanded, we get \[(2x + 7)(x - 4) = -31 + jx \Rightarrow 2x^2 - x - 8 = -31 + jx \Rightarrow 2x^2 - (j + 1)x + 23 = 0.\] For there to be exactly one real solution, the discriminant of this quadratic must be equal to 0. The discriminant is given by $b^2 - 4ac$ where $a=2$, $b=-(j+1)$, and $c=23$. Thus we have \[(-1 - j)^2 - 4(2)(23) = 0 \Rightarrow j^2 + 2j - 91 = 0.\] Factoring the quadratic gives $(j + 13)(j - 7) = 0$, so the possible values of $j$ are $j = \boxed{-13,7}$.
To find the values of $j$ that result in exactly one real solution for the equation, we need to determine when the quadratic equation formed by the expression $(2x + 7)(x - 4)$ is equal to the linear equation $-31 + jx$.
First, let's expand the quadratic expression:
$(2x + 7)(x - 4) = 2x(x) + 2x(-4) + 7(x) + 7(-4) = 2x^2 - 8x + 7x - 28 = 2x^2 - x - 28$
Now we set the quadratic equation equal to the linear equation:
$2x^2 - x - 28 = -31 + jx$
Rearranging terms, we get:
$2x^2 - x - 28 - jx + 31 = 0$
Combining like terms, we have:
$2x^2 - (1 + j)x + 3 = 0$
For this equation to have exactly one real solution, the discriminant must be equal to zero ($b^2 - 4ac = 0$).
Applying this condition in our equation, we have:
$(1 + j)^2 - 4(2)(3) = 0$
Expanding and simplifying:
$1 + 2j + j^2 - 24 = 0$
$j^2 + 2j - 23 = 0$
Now we can solve this quadratic equation for $j$. Using the quadratic formula $j = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we have:
$j = \frac{-2 \pm \sqrt{2^2 - 4(1)(-23)}}{2(1)}$
Simplifying further:
$j = \frac{-2 \pm \sqrt{4 + 92}}{2}$
$j = \frac{-2 \pm \sqrt{96}}{2}$
$j = \frac{-2 \pm 4\sqrt{6}}{2}$
$j = -1 \pm 2\sqrt{6}$
Therefore, the values of $j$ that result in the equation having exactly one real solution are $-1 + 2\sqrt{6}$ and $-1 - 2\sqrt{6}$.
In list form, the values of $j$ are $-1 + 2\sqrt{6}$, $-1 - 2\sqrt{6}$.