Equation of Tangents to a Circle from an External Point:
The equation of a tangent to a circle can be determined using the following steps:
1. Find the coordinates of the point of contact between the tangent and the circle.
2. Calculate the slope of the tangent line.
3. Use the point-slope form of a linear equation to find the equation of the tangent line.

Given Information:
The equation of the circle is x^2 + y^2 - 6x - 4y - 3 = 0. The point from which the tangents are drawn is (7, 4), which lies outside the circle.

Finding the Point of Contact:
To find the point of contact, we need to solve the simultaneous equations of the circle and the line passing through the given point (7, 4).

The equation of the line passing through (7, 4) can be written in the form y = mx + c, where m is the slope and c is the y-intercept. Since the line is a tangent to the circle, the point of contact lies on both the circle and the line.

Substituting y = mx + c into the equation of the circle, we get:
x^2 + (mx + c)^2 - 6x - 4(mx + c) - 3 = 0

Expanding and rearranging the terms, we obtain a quadratic equation in x:
(1 + m^2) x^2 + (2mc - 6 - 4m) x + (c^2 - 4c - 3) = 0

For the line to be a tangent, this quadratic equation must have only one real root. Therefore, the discriminant of the quadratic equation should be equal to zero.

The discriminant, b^2 - 4ac, is given by:
(2mc - 6 - 4m)^2 - 4(1 + m^2)(c^2 - 4c - 3) = 0

Simplifying the above equation, we get:
(4m^2c^2 - 24mc + 36m^2 + 36) - (4c^2 - 16c - 12m^2c^2 + 48mc + 36m^2) = 0

After canceling out like terms, we obtain:
12c^2 - 64c + 12m^2 = 0

This is a quadratic equation in c. Solving this equation will give us the value of c, which is the y-coordinate of the point of contact.

Calculating the Slope:
Once we have the value of c, we can substitute it back into the equation of the line (y = mx + c) to find the slope of the tangent.

Using the Point-Slope Form:
Finally, we can use the point-slope form of a linear equation (y - y1 = m(x - x1)) to find the equation of the tangent line. We substitute the slope and the coordinates of the point of contact into this equation to obtain the equation of the tangent.

Summary:
To find the equation of the tangents drawn to the given circle from the point (7, 4) lying outside the circle, follow