**2. The Quadratic Formula (Your Last Resort) **

All quadratics can be solved using this formula:

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

Let's solve the quadratic equation below.
\[ 2x^2 - 5x + 3 = 0 \]
\[ \Rightarrow a = 2, b = -5, c = 3\]

Substituting the values \( a = 2, b = -5, c = 3 \) into the formula, we get:
\[ x = \frac{{-(-5) \pm \sqrt{{(-5)^2 - 4(2)(3)}}}}{{2(2)}} \]

Simplifying further, we have:
\[ x = \frac{{5 \pm \sqrt{{25 - 24}}}}{{4}} \]
\[ x = \frac{{5 \pm \sqrt{1}}}{{4}} \]
\[ x = \frac{{5 \pm 1}}{{4}} \]
\[\Rightarrow x = \frac{3}{2} \text{ and } x = 1 \]

Therefore, the solutions to the equation are \( x = \frac{3}{2}\) and \( x = 1\)
**1. The Slope Formula **

Find the slope of the line passing through the points (2, 4) and (4, 10). To find the slope of the line passing through the points \( (2, 4) \) and \( (4, 10) \), we can use the formula:

\[ \text{slope} = \frac{{y_2 - y_1}}{{x_2 - x_1}} \]

Plugging in the coordinates, we have:
\[ \text{slope} = \frac{{10 - 4}}{{4 - 2}} = \frac{{6}}{{2}} = 3\]

Therefore, the slope of the line is 3.
**1. Slope with Variables **

If the line passes through the points \((3, a\)) and\( (6, 7a\)) with a slope of \(\frac{4}{3}\), find the value of \(a\).

To find the slope, we can use the formula:
\[
\text{Slope} = \frac{{y_2 - y_1}}{{x_2 - x_1}} = \frac{{7a - a}}{{6 - 3}} = 2a
\]

Now we are told the slope is \( \frac{4}{3}\), so we get:
\[ 2a = \frac{4}{3}
\]

Now we can solve for \(a\) by cross-multiplying:
\[
6a = 4 \quad \Rightarrow \quad a = \frac{4}{6} = \frac{2}{3}
\]

**1. Plugging values in \( y = mx + b \) **

\[ \text{Find the value of } a \text{ when the line } y = 4x + 6 \text{ passes through the point } (3, a). \]

To solve the problem, we can substitute the x-coordinate (3) and the y-coordinate (a) into \( y = 4x + 6 \).
So, we have:
\[ a = 4(3) + 6 \]
\[ a = 12 + 6 \]
\[ a = 18 \]

Therefore, the value of \(a\) is 18.

**1. The Slope-Intercept Form \( y = mx + b \) **

Given two points on a line, (2, 5) and (4, 9), find the slope and y-intercept.

To find the slope of the line passing through the points \( (2, 5) \) and \( (4, 9) \), we can use the formula:
\[ \text{slope} = \frac{{y_2 - y_1}}{{x_2 - x_1}} \]

Plugging in the coordinates, we have:
\[ \text{slope} = \frac{{9 - 5}}{{4 - 2}} = \frac{{4}}{{2}} = 2\]

Therefore, the slope of the line is 2 and the equation of the line is \( y = 2x + b \).

Next, we can plug in the coordinates (2, 5) into the equation \( y = 2x + b \) to find the value of b. Substituting \( x = 2 \) and \( y = 5 \):
\[ 5 = 2(2) + b \]
\[ 5 = 4 + b \]
\[ b = 5 - 4 = 1 \]

Therefore, the y-intercept is 1 and the equation of the line is \( y = 2x + 1 \).
**1. Parallel Lines \( y = mx + b \) **

A line passes through the points (2, 5) and (4, 9).

1) Determine the slope of this line.

2) Find the equation of a line parallel to this line that passes through the point (1, 10).

To find the slope of the line passing through (2, 5) and (4, 9), we can use the formula:
1) Determine the slope of this line.

2) Find the equation of a line parallel to this line that passes through the point (1, 10).

\[ m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \]

Substituting the given coordinates:

\[ m = \frac{{9 - 5}}{{4 - 2}} = \frac{4}{2} = 2 \]

Parallel lines have the same slope. So we can write the equation of the parallel line as
\[ \text{Equation of the parallel line: } y = 2x + b \]

To find the value of \( b \), we can substitute the coordinates \((1, 10)\) into the equation \( y = 2x + b \)
\[ 10 = 2(1) + b \]
\[ 10 = 2 + b \]
\[ 8 = b \]

Therefore, the value of \( b \) is 8 and the equation of the parallel line is \(y = 2x + 8 \).