Question

solve for n.
9n² + 20n + 4 = 0
write each solution as an integer, proper fraction, or improper fraction in simplest form. if there are multiple solutions, separate them with commas.

Explanation:

Step1: Identify coefficients

For the quadratic equation $9n^{2}+20n + 4=0$, we have $a = 9$, $b=20$, $c = 4$.

Step2: Use quadratic formula

The quadratic formula is $n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a = 9$, $b = 20$, $c=4$ into it:
First, calculate the discriminant $\Delta=b^{2}-4ac=(20)^{2}-4\times9\times4=400 - 144 = 256$.
Then, $n=\frac{-20\pm\sqrt{256}}{2\times9}=\frac{-20\pm16}{18}$.

Step3: Find two solutions

For the plus - sign: $n_1=\frac{-20 + 16}{18}=\frac{-4}{18}=-\frac{2}{9}$.
For the minus - sign: $n_2=\frac{-20-16}{18}=\frac{-36}{18}=- 2$.

Answer:

$-2,-\frac{2}{9}$