Collection of eight quadratic equations arranged in two columns for students to solve using the standard quadratic formula.

To solve each quadratic equation using the quadratic formula, we will follow these steps:

1. Identify the coefficients \(a\), \(b\), and \(c\) from the standard form of the quadratic equation \(ax^2 + bx + c = 0\).
2. Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
3. Substitute the values of \(a\), \(b\), and \(c\) into the formula.
4. Simplify the expression to find the solutions.

Let's solve each equation step by step.

Equation 1: \(3g^2 - 17g - 6 = 0\)

1. Identify coefficients: \(a = 3\), \(b = -17\), \(c = -6\).
2. Apply the quadratic formula:
\[
g = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(3)(-6)}}{2(3)}
\]
3. Simplify inside the square root:
\[
g = \frac{17 \pm \sqrt{289 + 72}}{6} = \frac{17 \pm \sqrt{361}}{6} = \frac{17 \pm 19}{6}
\]
4. Solve for the two possible values of \(g\):
\[
g = \frac{17 + 19}{6} = \frac{36}{6} = 6 \quad \text{and} \quad g = \frac{17 - 19}{6} = \frac{-2}{6} = -\frac{1}{3}
\]

Equation 2: \(4r^2 - 23r + 33 = 0\)

1. Identify coefficients: \(a = 4\), \(b = -23\), \(c = 33\).
2. Apply the quadratic formula:
\[
r = \frac{-(-23) \pm \sqrt{(-23)^2 - 4(4)(33)}}{2(4)}
\]
3. Simplify inside the square root:
\[
r = \frac{23 \pm \sqrt{529 - 528}}{8} = \frac{23 \pm \sqrt{1}}{8} = \frac{23 \pm 1}{8}
\]
4. Solve for the two possible values of \(r\):
\[
r = \frac{23 + 1}{8} = \frac{24}{8} = 3 \quad \text{and} \quad r = \frac{23 - 1}{8} = \frac{22}{8} = \frac{11}{4}
\]

Equation 3: \(24x^2 + 22x - 7 = 0\)

1. Identify coefficients: \(a = 24\), \(b = 22\), \(c = -7\).
2. Apply the quadratic formula:
\[
x = \frac{-22 \pm \sqrt{(22)^2 - 4(24)(-7)}}{2(24)}
\]
3. Simplify inside the square root:
\[
x = \frac{-22 \pm \sqrt{484 + 672}}{48} = \frac{-22 \pm \sqrt{1156}}{48} = \frac{-22 \pm 34}{48}
\]
4. Solve for the two possible values of \(x\):
\[
x = \frac{-22 + 34}{48} = \frac{12}{48} = \frac{1}{4} \quad \text{and} \quad x = \frac{-22 - 34}{48} = \frac{-56}{48} = -\frac{7}{6}
\]

Equation 4: \(2h^2 - 7h + 8 = 0\)

1. Identify coefficients: \(a = 2\), \(b = -7\), \(c = 8\).
2. Apply the quadratic formula:
\[
h = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(8)}}{2(2)}
\]
3. Simplify inside the square root:
\[
h = \frac{7 \pm \sqrt{49 - 64}}{4} = \frac{7 \pm \sqrt{-15}}{4}
\]
4. Since the discriminant is negative (\(\sqrt{-15}\)), there are no real solutions. The solutions are complex:
\[
h = \frac{7 \pm i\sqrt{15}}{4}
\]

Equation 5: \(20b^2 + 11b = 4\)

Rewrite in standard form: \(20b^2 + 11b - 4 = 0\).

1. Identify coefficients: \(a = 20\), \(b = 11\), \(c = -4\).
2. Apply the quadratic formula:
\[
b = \frac{-11 \pm \sqrt{(11)^2 - 4(20)(-4)}}{2(20)}
\]
3. Simplify inside the square root:
\[
b = \frac{-11 \pm \sqrt{121 + 320}}{40} = \frac{-11 \pm \sqrt{441}}{40} = \frac{-11 \pm 21}{40}
\]
4. Solve for the two possible values of \(b\):
\[
b = \frac{-11 + 21}{40} = \frac{10}{40} = \frac{1}{4} \quad \text{and} \quad b = \frac{-11 - 21}{40} = \frac{-32}{40} = -\frac{4}{5}
\]

Equation 6: \(12g^2 + 41g + 35 = 0\)

1. Identify coefficients: \(a = 12\), \(b = 41\), \(c = 35\).
2. Apply the quadratic formula:
\[
g = \frac{-41 \pm \sqrt{(41)^2 - 4(12)(35)}}{2(12)}
\]
3. Simplify inside the square root:
\[
g = \frac{-41 \pm \sqrt{1681 - 1680}}{24} = \frac{-41 \pm \sqrt{1}}{24} = \frac{-41 \pm 1}{24}
\]
4. Solve for the two possible values of \(g\):
\[
g = \frac{-41 + 1}{24} = \frac{-40}{24} = -\frac{5}{3} \quad \text{and} \quad g = \frac{-41 - 1}{24} = \frac{-42}{24} = -\frac{7}{4}
\]

Equation 7: \(5n^2 + 19n + 18 = 0\)

1. Identify coefficients: \(a = 5\), \(b = 19\), \(c = 18\).
2. Apply the quadratic formula:
\[
n = \frac{-19 \pm \sqrt{(19)^2 - 4(5)(18)}}{2(5)}
\]
3. Simplify inside the square root:
\[
n = \frac{-19 \pm \sqrt{361 - 360}}{10} = \frac{-19 \pm \sqrt{1}}{10} = \frac{-19 \pm 1}{10}
\]
4. Solve for the two possible values of \(n\):
\[
n = \frac{-19 + 1}{10} = \frac{-18}{10} = -\frac{9}{5} \quad \text{and} \quad n = \frac{-19 - 1}{10} = \frac{-20}{10} = -2
\]

Equation 8: \(6n^2 - 17n = 14\)

Rewrite in standard form: \(6n^2 - 17n - 14 = 0\).

1. Identify coefficients: \(a = 6\), \(b = -17\), \(c = -14\).
2. Apply the quadratic formula:
\[
n = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(6)(-14)}}{2(6)}
\]
3. Simplify inside the square root:
\[
n = \frac{17 \pm \sqrt{289 + 336}}{12} = \frac{17 \pm \sqrt{625}}{12} = \frac{17 \pm 25}{12}
\]
4. Solve for the two possible values of \(n\):
\[
n = \frac{17 + 25}{12} = \frac{42}{12} = \frac{7}{2} \quad \text{and} \quad n = \frac{17 - 25}{12} = \frac{-8}{12} = -\frac{2}{3}
\]

Final Answer:
\[
\boxed{
\begin{aligned}
&3g^2 - 17g - 6 = 0 \implies g = 6, -\frac{1}{3} \\
&4r^2 - 23r + 33 = 0 \implies r = 3, \frac{11}{4} \\
&24x^2 + 22x - 7 = 0 \implies x = \frac{1}{4}, -\frac{7}{6} \\
&2h^2 - 7h + 8 = 0 \implies h = \frac{7 \pm i\sqrt{15}}{4} \\
&20b^2 + 11b - 4 = 0 \implies b = \frac{1}{4}, -\frac{4}{5} \\
&12g^2 + 41g + 35 = 0 \implies g = -\frac{5}{3}, -\frac{7}{4} \\
&5n^2 + 19n + 18 = 0 \implies n = -\frac{9}{5}, -2 \\
&6n^2 - 17n - 14 = 0 \implies n = \frac{7}{2}, -\frac{2}{3}
\end{aligned}
}
\]