SOLUTION: solve for x 5x^2=-14x-8 Do I have to subtract 5x^2 from both sides giving me 0=-5x^2-14x-8? Then what do I do?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: solve for x 5x^2=-14x-8 Do I have to subtract 5x^2 from both sides giving me 0=-5x^2-14x-8? Then what do I do?      Log On


   



Question 490229: solve for x
5x^2=-14x-8
Do I have to subtract 5x^2 from both sides giving me 0=-5x^2-14x-8?
Then what do I do?

Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
5x^2=-14x-8
---
5x^2+14x+8 = 0
Factor:
5x^2+10x+4x+8 = 0
5x(x+2)+4(x+2) = 0
(x+2)(5x+4) = 0
---
x = -2 or x = -4/5
======================
Cheers,
Stan H.
===================

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

you can do this : 0=-5x%5E2-14x-8 and than use quadratic formula to solve for x
since you dealing with negative signs in this case, I suggest you do it this way: move both terms from the right to the left side
5x%5E2=-14x-8

5x%5E2%2B14x%2B8=0....now use quadratic formula to solve for x

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+...note a=5, b=14 and c=8

x+=+%28-14+%2B-+sqrt%28+14%5E2-4%2A5%2A8+%29%29%2F%282%2A5%29+

x+=+%28-14+%2B-+sqrt%28+196-160+%29%29%2F10+

x+=+%28-14+%2B-+sqrt%28+36+%29%29%2F10+

x+=+%28-14+%2B-+6%29%2F10+

solutions:
x+=+%28-14+%2B+6%29%2F10+

x+=+-8%2F10+

x+=+-4%2F5+

or
x+=+%28-14+-+6%29%2F10+

x+=+-20%2F10+

x+=+-2+

let's check it on a graph:
+graph%28+500%2C500%2C+-5%2C+5%2C+-10%2C+10%2C5x%5E2%2B14x%2B8%29+