Step 1. Simplify 10/3b to 10b/3
12/5(10b/3 + 15/4)
Step 2. Distribute
8b + 9
If the negative square root is found to be one of your solutions, then that is indicative of a pair of imaginary roots (the imaginary i). According to the conjugate rule, if you have one solution that is imaginary, you will have another but with the opposite sign. For example, if a solution to a quadratic is found to be 2 - i, then its conjugate, 2 + i is also a solution. They will ALWAYS go in pairs. Same thing with radical solutions. If one solution is found to be 
then
will also be a solution.
<h3>Answer:</h3>
Equation of the ellipse = 3x² + 5y² = 32
<h3>Step-by-step explanation:</h3>
<h2>Given:</h2>
- The centre of the ellipse is at the origin and the X axis is the major axis
- It passes through the points (-3, 1) and (2, -2)
<h2>To Find:</h2>
- The equation of the ellipse
<h2>Solution:</h2>
The equation of an ellipse is given by,

Given that the ellipse passes through the point (-3, 1)
Hence,

Cross multiplying we get,
- 9b² + a² = 1 ²× a²b²
- a²b² = 9b² + a²
Multiply by 4 on both sides,
- 4a²b² = 36b² + 4a²------(1)
Also by given the ellipse passes through the point (2, -2)
Substituting this,

Cross multiply,
- 4b² + 4a² = 1 × a²b²
- a²b² = 4b² + 4a²-------(2)
Subtracting equations 2 and 1,
- 3a²b² = 32b²
- 3a² = 32
- a² = 32/3----(3)
Substituting in 2,
- 32/3 × b² = 4b² + 4 × 32/3
- 32/3 b² = 4b² + 128/3
- 32/3 b² = (12b² + 128)/3
- 32b² = 12b² + 128
- 20b² = 128
- b² = 128/20 = 32/5
Substituting the values in the equation for ellipse,


Multiplying whole equation by 32 we get,
3x² + 5y² = 32
<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>