**Given information:**
- Initial velocity of the particle: 5i m/s
- Acceleration of the particle: 3i + 2j m/s²
- x-coordinate when the particle is at 84 m

**To find:**
- y-coordinate of the particle at the moment when its x-coordinate is 84 m

**Solution:**

Let's break down the problem into smaller steps.

**Step 1: Finding the x-coordinate equation:**

The problem states that the acceleration of the particle is constant. Therefore, we can use the following kinematic equation to find the x-coordinate of the particle as a function of time:

x = ut + 0.5at²

Given:
- Initial velocity (u) = 5i m/s
- Acceleration (a) = 3i + 2j m/s²
- t = ?
- x = 84 m

Since the acceleration is only in the x-direction, the y-component of acceleration (2j) does not affect the x-coordinate equation. Hence, we can ignore it.

Plugging in the values, we get:
84 = 5t + 0.5(3)t²

Simplifying the equation, we have:
84 = 5t + 1.5t²

**Step 2: Solving the quadratic equation:**

To solve the quadratic equation, we rearrange the equation to bring it to the standard form:
1.5t² + 5t - 84 = 0

Now, we can solve the quadratic equation either by factorization or by using the quadratic formula. Let's use the quadratic formula:

t = (-b ± √(b² - 4ac)) / 2a

Plugging in the values, we get:
t = (-5 ± √(5² - 4(1.5)(-84))) / (2(1.5))

Simplifying further, we get:
t = (-5 ± √(25 + 1008)) / 3

t = (-5 ± √1033) / 3

Since time cannot be negative, we consider the positive root of the equation:
t = (-5 + √1033) / 3

**Step 3: Finding the y-coordinate:**

Now that we have the value of t, we can find the y-coordinate using the equation of motion:

y = ut + 0.5at²

Given:
- Initial velocity (u) = 5i m/s
- Acceleration (a) = 3i + 2j m/s²
- t = (-5 + √1033) / 3

Plugging in the values, we get:
y = 5(-5 + √1033) / 3 + 0.5(3)(-5 + √1033)²

Simplifying the equation, we get the y-coordinate at the given x-coordinate of 84 m.

Therefore, the y-coordinate of the particle when its x-coordinate is 84 m is given by the equation above.