To solve the quadratic equation \( g^2 + 14g + 13 = 0 \), we can use the quadratic formula:
\[ g = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the given equation, we identify the coefficients as follows:
- \( a = 1 \)
- \( b = 14 \)
- \( c = 13 \)
Next, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 14^2 = 196 \] \[ 4ac = 4 \cdot 1 \cdot 13 = 52 \] \[ b^2 - 4ac = 196 - 52 = 144 \]
Now we plug the values into the quadratic formula:
\[ g = \frac{-14 \pm \sqrt{144}}{2 \cdot 1} \] \[ g = \frac{-14 \pm 12}{2} \]
This will give us two possible solutions for \( g \):
-
Using the \( + \) sign: \[ g = \frac{-14 + 12}{2} = \frac{-2}{2} = -1 \]
-
Using the \( - \) sign: \[ g = \frac{-14 - 12}{2} = \frac{-26}{2} = -13 \]
Thus, the solutions for \( g \) are:
\[ g = -1, -13 \]
So the final answer is:
\[ g = -1, -13 \]