(UPDATED POST) RC car bluetooth L293D motor driver

Can someone please help me. I cant properly control my rc car (im following a tutorial on YT). I did everything on the tutorial, i dont think im missing something but yeah i cant control it properly.
This is how i wire it and as you can see i didn't solder it yet because i want to make sure it works first.

The bluetooth rc contoller i used:

Schematic diagram:

is this connection possible?(i use the same l293d above)

The code:

#include <AFMotor.h>

AF_DCMotor motor1(1);  // Motor 1 connected to M1 port of L293D
AF_DCMotor motor2(2);  // Motor 2 connected to M2 port of L293D
AF_DCMotor motor3(3);  // Motor 3 connected to M3 port of L293D
AF_DCMotor motor4(4);  // Motor 4 connected to M4 port of L293D

int linearActuatorEnablePin = 11; // L298N Enable pin
int linearActuatorInput1 = 10;    // L298N Input 1
int linearActuatorInput2 = 9;     // L298N Input 2
int linearActuatorDelay = 3000;    // Delay before linear actuator movement (in milliseconds)
int linearActuatorMoveTime = 10000; // Time for linear actuator to extend and retract (in milliseconds)
unsigned long lastMoveTime = 0;

void setup() {
  Serial.begin(9600);  // Initialize serial communication
  motor1.setSpeed(255); // Set motor speed (0-255)
  motor2.setSpeed(255);
  motor3.setSpeed(255);
  motor4.setSpeed(255);
  pinMode(linearActuatorEnablePin, OUTPUT);
  pinMode(linearActuatorInput1, OUTPUT);
  pinMode(linearActuatorInput2, OUTPUT);
}

void loop() {
  if (Serial.available() > 0) {
    char command = Serial.read();
    executeCommand(command);
  }

  // Check if the car is not moving for 3 seconds
  if ((millis() - lastMoveTime) > linearActuatorDelay) {
    // Extend and retract the linear actuator after 10 seconds
    if ((millis() - lastMoveTime) > (linearActuatorDelay + linearActuatorMoveTime)) {
      moveLinearActuator();
      lastMoveTime = millis();
    }
  }
}

void executeCommand(char command) {
  switch (command) {
    case 'F':
      moveForward();
      break;
    case 'B':
      moveBackward();
      break;
    case 'L':
      turnLeft();
      break;
    case 'R':
      turnRight();
      break;
    case 'S':
      stopMoving();
      break;
  }
}

void moveForward() {
  motor1.setSpeed(255);
  motor2.setSpeed(255);
  motor3.setSpeed(255);
  motor4.setSpeed(255);
  motor1.run(FORWARD);
  motor2.run(FORWARD);
  motor3.run(FORWARD);
  motor4.run(FORWARD);
  lastMoveTime = millis();
}

void moveBackward() {
  motor1.setSpeed(255);
  motor2.setSpeed(255);
  motor3.setSpeed(255);
  motor4.setSpeed(255);
  motor1.run(BACKWARD);
  motor2.run(BACKWARD);
  motor3.run(BACKWARD);
  motor4.run(BACKWARD);
  lastMoveTime = millis();
}

void turnLeft() {
  motor1.setSpeed(255);
  motor2.setSpeed(255);
  motor3.setSpeed(255);
  motor4.setSpeed(255);
  motor1.run(BACKWARD);
  motor2.run(BACKWARD);
  motor3.run(FORWARD);
  motor4.run(FORWARD);
  lastMoveTime = millis();
}

void turnRight() {
  motor1.setSpeed(255);
  motor2.setSpeed(255);
  motor3.setSpeed(255);
  motor4.setSpeed(255);
  motor1.run(FORWARD);
  motor2.run(FORWARD);
  motor3.run(BACKWARD);
  motor4.run(BACKWARD);
  lastMoveTime = millis();
}

void stopMoving() {
  motor1.setSpeed(0);
  motor2.setSpeed(0);
  motor3.setSpeed(0);
  motor4.setSpeed(0);
}

void moveLinearActuator() {
  digitalWrite(linearActuatorEnablePin, HIGH);
  digitalWrite(linearActuatorInput1, HIGH);
  digitalWrite(linearActuatorInput2, LOW);
  delay(linearActuatorMoveTime);
  digitalWrite(linearActuatorEnablePin, LOW);
}

If battery life is important chose a H-bridge with MOSFET outputs. Your existing one loses 3V from the battery to the motor which is burnt up as heat.

what about L298n?

The voltage drop is inherent in the design of the L293. It has bipolar output transistors in a darling ton configuration. One for the High side one for the Low side. Each will drop about 0.7V per transistor so in a darling ton that is 0.7 * 1 or 1.4V to keep math simple and pick up wiring losses I round it up to 1.5V. See page 8 of the data sheet: https://www.ti.com/lit/ds/symlink/l293.pdf?ts=1700100271447&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FL293 This voltage drop is in each output pin. So with a motor connected it is ~2.8V less to the motor then the supply. Assuming a 6V battery the loss is about 50% which is burnt as heat in the part. Hopefully that helps explain what is happening. Using a MOSFET output I would expect the voltage drop to be maybe 0.1V or less.