The problem asks to solve for $x$ in the following two quadratic equations: a) $x^2 - 2x - 2 = 0$ b) $2x^2 + 3x - 4 = 0$ We will use the quadratic formula to solve for $x$.

1. Problem Description

The problem asks to solve for $x$ in the following two quadratic equations:

a) $x^2 - 2x - 2 = 0$

b) $2x^2 + 3x - 4 = 0$

We will use the quadratic formula to solve for $x$.

2. Solution Steps

a) For the quadratic equation $x^2 - 2x - 2 = 0$, we have $a = 1$, $b = -2$, and $c = -2$. The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the values of $a$, $b$, and $c$, we have:

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

$x = \frac{2 \pm \sqrt{4 + 8}}{2}$

$x = \frac{2 \pm \sqrt{12}}{2}$

$x = \frac{2 \pm 2\sqrt{3}}{2}$

$x = 1 \pm \sqrt{3}$

b) For the quadratic equation $2x^2 + 3x - 4 = 0$, we have $a = 2$, $b = 3$, and $c = -4$. Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the values of $a$, $b$, and $c$, we have:

$x = \frac{-3 \pm \sqrt{(3)^2 - 4(2)(-4)}}{2(2)}$

$x = \frac{-3 \pm \sqrt{9 + 32}}{4}$

$x = \frac{-3 \pm \sqrt{41}}{4}$

3. Final Answer

a) $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}$

b) $x = \frac{-3 + \sqrt{41}}{4}$ and $x = \frac{-3 - \sqrt{41}}{4}$