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