To find the possible values of n in the quadratic equation 2n² + 7n + 6 = 0, we can use the quadratic formula:
n = (-b ± √(b² – 4ac)) / 2a
In our equation, the coefficients are:
- a = 2
- b = 7
- c = 6
Now, we need to calculate the discriminant (b² – 4ac):
Discriminant = b² – 4ac = (7)² – 4(2)(6)
Discriminant = 49 – 48 = 1
Since the discriminant is greater than 0, we can conclude that there are two distinct real solutions for n. Now we can substitute the values of a, b, and the discriminant back into the quadratic formula:
n = (-7 ± √1) / (2 * 2)
n = (-7 ± 1) / 4
Now we can calculate the two possible values for n:
1st Solution:
n = (-7 + 1) / 4 = -6 / 4 = -1.5
2nd Solution:
n = (-7 – 1) / 4 = -8 / 4 = -2
So, the possible values of n for the equation 2n² + 7n + 6 = 0 are:
- n = -1.5
- n = -2