Algebra

First, let's combine all like terms for x and move all terms to one side. We have that

$x^{2}+12x-4=0$

Using the quadratic equation, we have that

$x=\frac{-12\pm \sqrt{12^{2}-4\cdot 1(-4)}}{2\cdot 1}$

$x=2\sqrt{10}-6\\ x=-2\sqrt{10}-6$

Now, we can use a handy trick to solve this problem. We have that

$(2\sqrt{10}-6)^2+(-2\sqrt{10}-6)^2$

Expanding and distributing in our values, we have

$-24\sqrt{10}+76+4\left(\sqrt{10}\right)^{2}+12\sqrt{10}-6\left(-2\sqrt{10}-6\right)$

All the radicals basically cancel out. In the end, we are left with 152. 

So 152 is our answer. 

Thanks! :)

Simplifying we get $x^2 + 12x - 4 = 0$. Using the quadratic formula we find the roots to be $-6 + 2\sqrt{10}$ and $-6 - 2\sqrt{10}$. Taking their squares, we get $36 + 4\sqrt{10} + 40$ and $36 - 4\sqrt{10} + 40$. Adding them together, we get $\boxed{152}$

Let the roots be a,b

Product of the roots  = ab =  -4   →  2ab = -8

Sum of the roots =  a + b = -12     square both sides

a^2 + b^2 + 2ab = 144

a^2 + b^2 - 8 =144

a^2 + b^2 =  152