To solve the equation \( 26n^2 + 41n + 20 = 6n^2 \), we first move all terms to one side of the equation to set it to zero:
\[ 26n^2 + 41n + 20 - 6n^2 = 0 \]
Combine the \( n^2 \) terms:
\[ (26n^2 - 6n^2) + 41n + 20 = 0 \] \[ 20n^2 + 41n + 20 = 0 \]
Now we can use the quadratic formula, which is given by:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \( a = 20 \), \( b = 41 \), and \( c = 20 \).
Now, let's calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 41^2 = 1681 \] \[ 4ac = 4 \cdot 20 \cdot 20 = 1600 \] \[ b^2 - 4ac = 1681 - 1600 = 81 \]
Now, we substitute the values into the quadratic formula:
\[ n = \frac{-41 \pm \sqrt{81}}{2 \cdot 20} \] \[ n = \frac{-41 \pm 9}{40} \]
Now we calculate the two possible solutions:
- For the plus case:
\[ n = \frac{-41 + 9}{40} = \frac{-32}{40} = \frac{-8}{10} = \frac{-4}{5} \]
- For the minus case:
\[ n = \frac{-41 - 9}{40} = \frac{-50}{40} = \frac{-5}{4} \]
Thus, the solutions to the equation \( 26n^2 + 41n + 20 = 6n^2 \) are:
\[ n = \frac{-4}{5} \quad \text{and} \quad n = \frac{-5}{4} \]