Use the quadratic formula to solve. Express your answer in simplest form.

26, n, squared, plus, 41, n, plus, 20, equals, 6, n, squared
26n
2
+41n+20=
6n
2

1 answer

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:

  1. For the plus case:

\[ n = \frac{-41 + 9}{40} = \frac{-32}{40} = \frac{-8}{10} = \frac{-4}{5} \]

  1. 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:

\[ n = \frac{-4}{5} \quad \text{and} \quad n = \frac{-5}{4} \]