3x^2 - 4x + 1 =0
This is what I did.
-(-4) = -4^2 - 4(3)(1) / 2(3)
4 = 16 - 12 / 6
4 = 4/6 = 1/3
Another answer is 1. How do they get 1.

x(x+6) = -11 - x
x^2 + 6x + x = -11
-6 = 6^2 - (4)(1)(1) / 2(1)
6 = 36 - 4 / 2
6 = 32 / 2
What do I do next?

2. I'm not sure why there are extra equal signs ....

$3x^{2} -4 x + 1 = 0$

$x = \frac{-(-4) \: {\color{red} \pm}\: \sqrt{(-4)^{2} - 4(3)(1)}}{2(3)}$

$x = \frac{4 \:{\color{red} \pm}\: \sqrt{16 - 12}}{6}$

$x = \frac{4 \:{\color{red} \pm}\: \sqrt{4}}{6} = \frac{4 \pm 2}{6}$

$x = \frac{4+2}{6} = 1 \quad \text{and} \quad x = \frac{4 - 2}{6} = \frac{1}{3}$

3. How do you do the second problem?

4. $x(x+6) = -11 - x$
$x^{2} + 6x = -11 - x$
$x^{2} + 7x + 11 = 0$