1) To solve the equation 4x^2-12x+9=0, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2+bx+c=0, the solutions for x are given by:

x = (-b ± √(b^2-4ac)) / (2a)

In this case, a = 4, b = -12, and c = 9. Plugging these values into the quadratic formula, we get:

x = (-(-12) ± √((-12)^2-4(4)(9))) / (2(4))
x = (12 ± √(144-144)) / 8
x = (12 ± √0) / 8
x = (12 ± 0) / 8
x = 12 / 8
x = 3/2

Therefore, the solution to the equation 4x^2-12x+9=0 is x = 3/2.

2) To solve the equation 3x^2+5x+2=0, we can again use the quadratic formula. In this case, a = 3, b = 5, and c = 2. Plugging these values into the quadratic formula, we get:

x = (-5 ± √(5^2-4(3)(2))) / (2(3))
x = (-5 ± √(25-24)) / 6
x = (-5 ± √1) / 6
x = (-5 ± 1) / 6

This gives us two possible solutions:
x = (-5 + 1) / 6 = -4 / 6 = -2 / 3
x = (-5 - 1) / 6 = -6 / 6 = -1

Therefore, the solutions to the equation 3x^2+5x+2=0 are x = -2/3 and x = -1.

3) To solve the equation x^2-7x+12=0, we can factorize the quadratic expression. We are looking for two numbers that multiply to 12 and add up to -7. The numbers -3 and -4 satisfy these conditions, so we can rewrite the equation as:

(x - 3)(x - 4) = 0

Setting each factor equal to zero, we get:

x - 3 = 0 or x - 4 = 0

Solving for x in each equation, we find:

x = 3 or x = 4

Therefore, the solutions to the equation x^2-7x+12=0 are x = 3 and x = 4.

4) To solve the equation x^2-2x-3=0, we can again factorize the quadratic expression. We are looking for two numbers that multiply to -3 and add up to -2. The numbers -3 and 1 satisfy these conditions, so we can rewrite the equation as:

(x - 3)(x + 1) = 0

Setting each factor equal to zero, we get:

x - 3 = 0 or x + 1 = 0

Solving for x in each equation, we find:

x = 3 or x = -1

Therefore, the solutions to the equation x^2-2x-3=0 are x = 3 and x = -1.