SOLUTION: Question: Solve 2x^2+3x=20 (a) [4, -2/5] (b) [4, -5/2] (c) [-4, 5/2] (d) [-4, 2/5] Thanks!

Algebra ->  College  -> Linear Algebra -> SOLUTION: Question: Solve 2x^2+3x=20 (a) [4, -2/5] (b) [4, -5/2] (c) [-4, 5/2] (d) [-4, 2/5] Thanks!      Log On


   

Question 30349: Question:
Solve
2x^2+3x=20
(a) [4, -2/5]
(b) [4, -5/2]
(c) [-4, 5/2]
(d) [-4, 2/5]
Thanks!
Answer by mukhopadhyay(490) About Me  (Show Source):
You can put this solution on YOUR website!
2x^2 + 3x = 20
=> 2x^2 + 3x - 20 = 0
=> 2x^2 + 8x - 5x - 20 = 0
=> 2x(x+4) - 5(x+4) = 0
=> (x+4)(2x-5) = 0
=> x = -4 or x = 5/2