SOLUTION: Solve the following Quadratic equation 4x+x(x-3)=5 Please help me figure this ine out.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve the following Quadratic equation 4x+x(x-3)=5 Please help me figure this ine out.      Log On


   

Question 132944: Solve the following Quadratic equation
4x+x(x-3)=5 Please help me figure this ine out.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
4x%2Bx%28x-3%29=5

In order to solve a quadratic, it needs to be in standard form (ax%5E2%2Bbx%2Bc=0, so:

Distribute the x:
4x%2Bx%5E2-3x=5

Collect like terms:
x%2Bx%5E2=5

Add -5 to both sides:
x%2Bx%5E2-5=0

And rearrange the terms in decending degree order:
x%5E2%2Bx-5=0

This quadratic does not factor to integer factors, so let's try completing the square.

Move the constant term back to the right side:
x%5E2%2Bx=5

Divide the coefficient on the first degree term by 2: 1%2F2, square it, and add the result to both sides:
x%5E2%2Bx%2B1%2F4=21%2F4

Now the left side is a perfect square, so factor it:
%28x%2B1%2F2%29%5E2=21%2F4

Take the square root of both sides:
x%2B1%2F2=sqrt%2821%2F4%29 or x%2B1%2F2=-sqrt%2821%2F4%29

x=-1%2F2%2B-sqrt%2821%2F4%29 => x=%28-1%2B-sqrt%2821%29%29%2F2