SOLUTION: I had to solve a quadratic equation by factoring and I recived it back as incorect. I dont know what I did wrong...could someone please explain this to me and show the correct step

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: I had to solve a quadratic equation by factoring and I recived it back as incorect. I dont know what I did wrong...could someone please explain this to me and show the correct step      Log On


   

Question 112700: I had to solve a quadratic equation by factoring and I recived it back as incorect. I dont know what I did wrong...could someone please explain this to me and show the correct steps if possible. I did not get 2 solutions for x and I cant figure out how to do this.
Thanks so much.
3x^2+11x-20=0
3x^2+15x-4x-20=0
3x(x+15)-4(x+5)=0
(3x-4)(x+5=0)
Found 2 solutions by jim_thompson5910, MathLover1:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%283x-4%29%28x%2B5%29=0 I'm going to start where you left off


3x-4=0 or x%2B5=0 Set each factor equal to zero


x=4%2F3 or x=-5 Solve for x in each case

So our solution is x=4%2F3 or x=-5

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2%2B11x-20=0
3x%5E2%2B15x-4x-20=0
3x%28x%2B15%29-4%28x%2B5%29=0
%283x-4%29%28x%2B5%29=0....well done up to here...
now, the product above will be equal to 0 if one or both factors are equal to 0
let find out what are values for x%5B1%2C2%5D:
3x+-+4+=+0
3x+=+4
x+=+4%2F3
if x+=+1.33........first solution
x+%2B+5+=+0
if x+=+-5.........second solution