To find the solutions of the equation 2x² + 3x + 5 = 0, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / (2a)
In this equation:
- a = 2
- b = 3
- c = 5
First, we’ll calculate the discriminant (b² – 4ac):
Discriminant = (3)² – 4(2)(5)
This results in:
Discriminant = 9 – 40 = -31
Since the discriminant is negative, this indicates that there are no real solutions to this equation, but there are two complex solutions. We proceed with the quadratic formula:
x = (-3 ± √(-31)) / (4)
Now, we can simplify the square root of the negative number:
√(-31) = i√31 (where i is the imaginary unit)
Substituting back into the formula gives:
x = (-3 ± i√31) / 4
Thus, the two solutions can be expressed as:
- x₁ = (-3 + i√31) / 4
- x₂ = (-3 – i√31) / 4
So, the two complex solutions to the equation 2x² + 3x + 5 = 0 are:
- Solution 1: (-3 + i√31) / 4
- Solution 2: (-3 – i√31) / 4
These solutions can be written as (-3 + i√31) / 4, (-3 – i√31) / 4 separated by a comma.