Elementary algebra worksheet focusing on solving quadratic equations using the quadratic formula, featuring 20 problems with various variables and coefficients.

To solve quadratic equations using the quadratic formula, we use the formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where the quadratic equation is in the standard form \( ax^2 + bx + c = 0 \).

Let's solve each equation step by step.

---

1. \( 3n^2 - 5n - 8 = 0 \)

Here, \( a = 3 \), \( b = -5 \), and \( c = -8 \).

\[
n = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-8)}}{2(3)}
\]

\[
n = \frac{5 \pm \sqrt{25 + 96}}{6}
\]

\[
n = \frac{5 \pm \sqrt{121}}{6}
\]

\[
n = \frac{5 \pm 11}{6}
\]

This gives two solutions:

\[
n = \frac{5 + 11}{6} = \frac{16}{6} = \frac{8}{3}
\]

\[
n = \frac{5 - 11}{6} = \frac{-6}{6} = -1
\]

So, the solutions are:

\[
n = \frac{8}{3}, \quad n = -1
\]

---

2. \( x^2 + 10x + 21 = 0 \)

Here, \( a = 1 \), \( b = 10 \), and \( c = 21 \).

\[
x = \frac{-10 \pm \sqrt{10^2 - 4(1)(21)}}{2(1)}
\]

\[
x = \frac{-10 \pm \sqrt{100 - 84}}{2}
\]

\[
x = \frac{-10 \pm \sqrt{16}}{2}
\]

\[
x = \frac{-10 \pm 4}{2}
\]

This gives two solutions:

\[
x = \frac{-10 + 4}{2} = \frac{-6}{2} = -3
\]

\[
x = \frac{-10 - 4}{2} = \frac{-14}{2} = -7
\]

So, the solutions are:

\[
x = -3, \quad x = -7
\]

---

3. \( 10x^2 - 9x + 6 = 0 \)

Here, \( a = 10 \), \( b = -9 \), and \( c = 6 \).

\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(10)(6)}}{2(10)}
\]

\[
x = \frac{9 \pm \sqrt{81 - 240}}{20}
\]

\[
x = \frac{9 \pm \sqrt{-159}}{20}
\]

Since the discriminant (\( -159 \)) is negative, there are no real solutions. The solutions are complex:

\[
x = \frac{9 \pm i\sqrt{159}}{20}
\]

So, the solutions are:

\[
x = \frac{9 + i\sqrt{159}}{20}, \quad x = \frac{9 - i\sqrt{159}}{20}
\]

---

4. \( p^2 - 9 = 0 \)

This can be rewritten as:

\[
p^2 = 9
\]

Taking the square root of both sides:

\[
p = \pm 3
\]

So, the solutions are:

\[
p = 3, \quad p = -3
\]

---

5. \( 6x^2 - 12x + 1 = 0 \)

Here, \( a = 6 \), \( b = -12 \), and \( c = 1 \).

\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(6)(1)}}{2(6)}
\]

\[
x = \frac{12 \pm \sqrt{144 - 24}}{12}
\]

\[
x = \frac{12 \pm \sqrt{120}}{12}
\]

\[
x = \frac{12 \pm 2\sqrt{30}}{12}
\]

\[
x = \frac{6 \pm \sqrt{30}}{6}
\]

So, the solutions are:

\[
x = \frac{6 + \sqrt{30}}{6}, \quad x = \frac{6 - \sqrt{30}}{6}
\]

---

6. \( 6n^2 - 11 = 0 \)

This can be rewritten as:

\[
6n^2 = 11
\]

\[
n^2 = \frac{11}{6}
\]

Taking the square root of both sides:

\[
n = \pm \sqrt{\frac{11}{6}}
\]

So, the solutions are:

\[
n = \sqrt{\frac{11}{6}}, \quad n = -\sqrt{\frac{11}{6}}
\]

---

7. \( 2n^2 + 5n - 9 = 0 \)

Here, \( a = 2 \), \( b = 5 \), and \( c = -9 \).

\[
n = \frac{-5 \pm \sqrt{5^2 - 4(2)(-9)}}{2(2)}
\]

\[
n = \frac{-5 \pm \sqrt{25 + 72}}{4}
\]

\[
n = \frac{-5 \pm \sqrt{97}}{4}
\]

So, the solutions are:

\[
n = \frac{-5 + \sqrt{97}}{4}, \quad n = \frac{-5 - \sqrt{97}}{4}
\]

---

8. \( 3x^2 - 6x - 23 = 0 \)

Here, \( a = 3 \), \( b = -6 \), and \( c = -23 \).

\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(-23)}}{2(3)}
\]

\[
x = \frac{6 \pm \sqrt{36 + 276}}{6}
\]

\[
x = \frac{6 \pm \sqrt{312}}{6}
\]

\[
x = \frac{6 \pm 2\sqrt{78}}{6}
\]

\[
x = \frac{3 \pm \sqrt{78}}{3}
\]

So, the solutions are:

\[
x = \frac{3 + \sqrt{78}}{3}, \quad x = \frac{3 - \sqrt{78}}{3}
\]

---

Final Answer

\[
\boxed{
\begin{aligned}
1. & \quad n = \frac{8}{3}, \, n = -1 \\
2. & \quad x = -3, \, x = -7 \\
3. & \quad x = \frac{9 + i\sqrt{159}}{20}, \, x = \frac{9 - i\sqrt{159}}{20} \\
4. & \quad p = 3, \, p = -3 \\
5. & \quad x = \frac{6 + \sqrt{30}}{6}, \, x = \frac{6 - \sqrt{30}}{6} \\
6. & \quad n = \sqrt{\frac{11}{6}}, \, n = -\sqrt{\frac{11}{6}} \\
7. & \quad n = \frac{-5 + \sqrt{97}}{4}, \, n = \frac{-5 - \sqrt{97}}{4} \\
8. & \quad x = \frac{3 + \sqrt{78}}{3}, \, x = \frac{3 - \sqrt{78}}{3}
\end{aligned}
}
\]