SOLUTION: Is this even considered to be a quadratic equation?? I do not understand by what they mean to solve this.. Do they mean distribute and then solve? But how? Please help! I am very f

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Is this even considered to be a quadratic equation?? I do not understand by what they mean to solve this.. Do they mean distribute and then solve? But how? Please help! I am very f      Log On


   

Question 559574: Is this even considered to be a quadratic equation?? I do not understand by what they mean to solve this.. Do they mean distribute and then solve? But how? Please help! I am very frustrated! Thanks!
(x-3)(5x+2)= -13(x-3)
Found 2 solutions by ankor@dixie-net.com, richard1234:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
(x-3)(5x+2) = -13(x-3)
Hey! we don't have to do a lot here, divide both sides by (x-3), results
5x + 2 = -13
5x = -13 -2
5x = -15
x = %28-15%29%2F5
x = -3
;
:
Check it:
(-3-3)(5(-3)+2) = -13(-3-3)
-6 * -13 = -13 * -6

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
The other tutor missed the solution x = 3 (since he divided by x-3, assuming it wasn't equal to 0).

One way to do it is to claim that x = 3 works (which it does). If x is *not* equal to 3, then you may divide both sides by x-3 to obtain the other solution, x = -3.