To find the solution set for the quadratic equation 6x² + 13x + 5 = 0, we need to use the quadratic formula:
x = (-b ± √(b² – 4ac)) / 2a
In our equation, the coefficients are:
- a = 6
- b = 13
- c = 5
First, we need to calculate the discriminant, which is b² – 4ac:
D = 13² – 4(6)(5) = 169 – 120 = 49
The discriminant is positive, indicating that there are two distinct real roots. Now we can substitute the values into the quadratic formula:
x = (−13 ± √49) / (2 6)
This simplifies to:
x = (−13 ± 7) / 12
Now, we calculate the two possible values for x:
- x₁ = (−13 + 7) / 12 = -6 / 12 = -1/2
- x₂ = (−13 − 7) / 12 = -20 / 12 = -5/3
Thus, the solution set for the equation 6x² + 13x + 5 = 0 is:
{ -1/2, -5/3 }