why SAR ADC charge sign is negative
I think the polarity taken to define \$V_+\$ in your slide is wrong. Let me show the full development of the SAR ADC equations.



simulate this circuit &ndash; Schematic created using CircuitLab
On \$0&lt;t&lt;t_1\$, SW1 is closed and SW4, SW7, SW10 and SW13 closed. The rest remain open as shown:



simulate this circuit
Let \$V_3\$, \$V_2\$, \$V_1\$, \$V_0\$ be the voltage between V+ and ground on \$C_3\$, \$C_2\$, \$C_1\$, \$C_0\$ capacitors repectively, the total charge in the system will be:
$$ Q_T = V_3 \cdot C_3 + V_2 \cdot C_2 + V_1 \cdot C_1 + V_0 \cdot C_0 $$
$$= \left(0-V_{in}\right) \cdot C_3 + \left(0-V_{in}\right) \cdot C_2 + \left(0-V_{in}\right) \cdot C_1 + V_0 \cdot \left(0-V_{in}\right) $$
$$= -V_{in} \cdot \left(C_3 + C_2 + C_1+C_0\right) $$
$$= -V_{in} \cdot \left(4C + 2C + C + C\right) $$
$$= -V_{in} \cdot 8C $$

Now, in the second phase (\$t_1 &lt; t &lt; t_2\$), SW1 is opened, and only switches SW2, SW6, SW8 and SW12 are closed:



simulate this circuit
The total charge of the system now will be:
$$Q^\prime_T = V_3 \cdot C_3 + V_2 \cdot C_2 + V_1 \cdot C_1 + V_0 \cdot C_0 $$
$$= \left(V_+ - V_{ref}\right) \cdot C_3 + \left(V_+ - 0 \right) \cdot C_2 +  \left(V_+ - V_{ref}\right) \cdot C_1 + \left(V_+ -0\right) \cdot C_0$$
$$= \left(C_3 + C_2 + C_1 + C_0\right) \cdot V_+ - \left(C_3 + C_1 \right) V_{ref}$$
$$= \left(4C + 2C + C +C\right) \cdot V_+ - \left(4C + C \right) \cdot V_{ref}$$
$$= 8C \cdot V_+ - 5C \cdot V_{ref}$$

Since the charge must keep constant before and after the switches change position:
$$ Q_T = Q^\prime_T$$
$$ -V_{in} \cdot 8C =  8C \cdot V_+ - 5C \cdot V_{ref}$$
$$ \boxed{V_+ = -V_{in} + \frac{5}{8} V_{ref}}$$

As you can see in the previous developed expressions, capacitor voltages are defined as:

Positive value: the common node where SW1 is
Negative value: the node connected to the 3 switches below each cap.

They will only match with the ones in your slides if \$V_+\$ was defined as:

Positive value: the node connected to the 3 switches below each cap.
Negative value: the common node where SW1 is

However, this definition would lead to an incorrect value of the voltage at the positive node of the operational amplifier. For this reason, I think that the whole expression for the \$V_+\$ on your slide must be multiplied by \$-1\$ in order to define properly \$V_+\$ as the voltage between the positive terminal of the operational amplifier and ground.