The sum of the squares of all the roots of the equation $$x^{2}+|2x-3|-4=0$$, is

We have |2x-3| so the sign changes at x = 3/2

Let's take case 1: $$x\ge\dfrac{3}{2}$$

In this case: $$|2x-3| = 2x-3$$.

Hence, the equation will be $$x^{2} + (2x - 3) - 4 = 0$$

Or, $$x^{2} + 2x - 7 = 0$$

Solving using the discriminant formula i.e. $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

$$x = \dfrac{-2 \pm \sqrt{2^2 - 4(1)(-7)}}{2(1)}$$

$$x = \dfrac{-2 \pm \sqrt{4 + 28}}{2}$$

$$x = \dfrac{-2 \pm \sqrt{32}}{2}$$

$$x = \dfrac{-2 \pm 4\sqrt{2}}{2}$$

$$x = -1 \pm 2\sqrt{2}$$

Now, let's check if the roots satisfy the initial assumption or not. 

$$x_1 = -1 + 2\sqrt{2} \approx -1 + 2(1.414) = 1.828$$. This is greater than 1.5, so it is a valid root.

$$x_2 = -1 - 2\sqrt{2} \approx -1 - 2.828 = -3.828$$. This is less than 1.5, so we reject it.

Let's take case 1:$$x<\dfrac{3}{2}$$

Here: $$|2x - 3| = -(2x - 3) = -2x + 3$$

Hence, the quadratic equation will be $$x^{2} + (-2x + 3) - 4 = 0$$

$$x^{2} - 2x - 1 = 0$$

Again, use the quadratic formula:

$$x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$$

$$x = \dfrac{2 \pm \sqrt{4 + 4}}{2}$$

$$x = \dfrac{2 \pm \sqrt{8}}{2}$$

$$x = \dfrac{2 \pm 2\sqrt{2}}{2}$$

$$x = 1 \pm \sqrt{2}$$

Again, we need to check if the roots satisfy the range or not. 

$$x_3 = 1 + \sqrt{2} \approx 1 + 1.414 = 2.414$$. This is greater than $$1.5$$, so we reject it.

$$x_4 = 1 - \sqrt{2} \approx 1 - 1.414 = -0.414$$ This is less than $$1.5$$, so it is a valid root.

Now, calculating the sum of the squares:

$$\alpha = -1 + 2\sqrt{2}$$ and $$\beta = 1 - \sqrt{2}$$

$$\alpha^{2} = (-1 + 2\sqrt{2})^{2} = 1 - 4\sqrt{2} + 4(2) = 1 - 4\sqrt{2} + 8 = 9 - 4\sqrt{2}$$

$$\beta^{2} = (1 - \sqrt{2})^{2} = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$$

Adding them: $$\alpha^{2} + \beta^{2} = (9 - 4\sqrt{2}) + (3 - 2\sqrt{2})$$

$$\alpha^2+\beta^2=12-6\sqrt{2}\ =\ 6\left(2-\sqrt{2}\right)$$