SOLUTION: Hello there: There are only four questions that I need help with. Thank you so much in advance, I really appreciate all of your help. 1. Solve (2x – 3)(x + 2) = 0 2. 6

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Hello there: There are only four questions that I need help with. Thank you so much in advance, I really appreciate all of your help. 1. Solve (2x – 3)(x + 2) = 0 2. 6      Log On


   



Question 426538: Hello there:
There are only four questions that I need help with.
Thank you so much in advance,
I really appreciate all of your help.
1. Solve (2x – 3)(x + 2) = 0
2. 6 = a^2 – 5a
3. 12x + 4 = -9x^2
4. 4y^2 = 25

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
1. Solve (2x – 3)(x + 2) = 0
Set each expression inside parentheses to zero and solve for x:
2x-3 = 0
2x = 3
x = 3/2
.
x+2 = 0
x = -2
.
Therefore,the solution is:
x = {-2, 3/2}
.
2. 6 = a^2 – 5a
move all terms to one side:
0 = a^2 – 5a - 6
factoring the right side:
0 = (a-6)(a+1)
a = {-1, 6}
.
3. 12x + 4 = -9x^2
9x^2 + 12x + 4 = 0
factoring the left side:
(3x+2)(3x+2) = 0
x = {-2/3}
.
4. 4y^2 = 25
dividing both sides by 4:
y^2 = 25/4
take the square root of both sides
y = 5/2