use factoring and the zero-product property to solve

4a^2-11a+6=0

9r^2-30r+21=-4

for 1 (4x-3)(x-2)

for 2 (3x-3)(3x-7)

can you explain how you got that? i'm completely lost

To solve the quadratic equation 4a^2 - 11a + 6 = 0 using factoring and the zero-product property, follow these steps:

Step 1: Rewrite the equation in the form of a quadratic expression equal to zero.
4a^2 - 11a + 6 = 0

Step 2: Factor the quadratic expression.
(2a - 1)(2a - 6) = 0

Step 3: Apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero.
2a - 1 = 0 or 2a - 6 = 0

Step 4: Solve each equation separately.
For 2a - 1 = 0:
Add 1 to both sides: 2a = 1
Divide both sides by 2: a = 1/2

For 2a - 6 = 0:
Add 6 to both sides: 2a = 6
Divide both sides by 2: a = 3

The solutions to the equation 4a^2 - 11a + 6 = 0 are a = 1/2 and a = 3.

Now, let's solve the second quadratic equation 9r^2 - 30r + 21 = -4 using the same method:

Step 1: Rewrite the equation in the form of a quadratic expression equal to zero.
9r^2 - 30r + 21 + 4 = 0
9r^2 - 30r + 25 = 0

Step 2: Factor the quadratic expression.
(3r - 5)(3r - 5) = 0

Step 3: Apply the zero-product property.
3r - 5 = 0

Step 4: Solve the equation.
Add 5 to both sides: 3r = 5
Divide both sides by 3: r = 5/3

The solution to the equation 9r^2 - 30r + 21 = -4 is r = 5/3.