That is English

You are absolutely right, Taydey! My apologies, that was indeed an English Cloze Test. Thanks for the correction.

Let's get back to the math problems from your image. Here are the solutions for questions 18, 19, and 20:

18. Solve for $x$ in $3^{2x+1} + 3^{x} - 12 = 0$.

Step 1: Rewrite the first term using exponent rules: $3^{2x+1} = 3^{2x} \cdot 3^1 = 3 \cdot (3^x)^2$.
The equation becomes:

$$3 \cdot (3^x)^2 + 3^x - 12 = 0$$

Step 2: Let $y = 3^x$. Substitute $y$ into the equation to form a quadratic equation.

$$3y^2 + y - 12 = 0$$

Step 3: Solve the quadratic equation for $y$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a=3$, $b=1$, $c=-12$.

$$y = \frac{-1 \pm \sqrt{1^2 - 4(3)(-12)}}{2(3)}$$ $$y = \frac{-1 \pm \sqrt{1 + 144}}{6}$$ $$y = \frac{-1 \pm \sqrt{145}}{6}$$

Step 4: Calculate the two possible values for $y$.

$$\sqrt{145} \approx 12.0416$$ $$y_1 = \frac{-1 + 12.0416}{6} = \frac{11.0416}{6} \approx 1.8403$$ $$y_2 = \frac{-1 - 12.0416}{6} = \frac{-13.0416}{6} \approx -2.1736$$

Step 5: Substitute back $y = 3^x$.
For $y_1$:

$$3^x = 1.8403$$

Take the logarithm of both sides (base 10 or natural log). Using base 10:

$$x \log 3 = \log 1.8403$$ $$x = \frac{\log 1.8403}{\log 3}$$ $$x = \frac{0.2649}{0.4771}$$ $$x \approx 0.5552$$

For $y_2$:

$$3^x = -2.1736$$

Since $3^x$ must always be positive for real values of $x$, this solution is not valid.
Therefore, the only valid solution for $x$ is:

$$x = \boxed{0.5552}$$

19. The sum of the first $n$ terms of an arithmetic progression is given by $S_n = 2n^2 - 6n$.