SOLUTION: (x-5)(x+2)=6. Thank you for your time, Don...

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Question 223846: (x-5)(x+2)=6. Thank you for your time, Don...
Found 2 solutions by checkley77, drj:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
(x-5)(x+2)=6
x^2-3x-10=6
x^2-3x-10-6=0
x^2-3x-16=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x=(3+-sqrt[-3^2-4*1*-16])/281
x=(3+-sqrt9+64])/2
x=(3+-sqrt73)/2
x=(3+-8.544)/2
x=(3+8.544)/2
x=11.544/2
x=5.772 ans.
x=(3-8.544)/2
x=(-5.544)/2
x=-2.772 ans.

Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-5%29%28x%2B2%29=6

Step 1. Multiply out the terms to get rid of the parenthesis and obtain a quadratic equation

%28x-5%29%28x%2B2%29=x%5E2%2B2x-5x-10=6

x%5E2-3x-10=6

Add -6 to both sides of the equation

x%5E2-3x-10%2B%28-6%29=6%2B%28-6%29

x%5E2-3x-16=0

Step 2. To solve, use the quadratic formula given as

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

where a=1, b=-3, and c=-16

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-3x%2B-16+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-3%29%5E2-4%2A1%2A-16=73.

Discriminant d=73 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--3%2B-sqrt%28+73+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-3%29%2Bsqrt%28+73+%29%29%2F2%5C1+=+5.77200187265877
x%5B2%5D+=+%28-%28-3%29-sqrt%28+73+%29%29%2F2%5C1+=+-2.77200187265877

Quadratic expression 1x%5E2%2B-3x%2B-16 can be factored:
1x%5E2%2B-3x%2B-16+=+1%28x-5.77200187265877%29%2A%28x--2.77200187265877%29
Again, the answer is: 5.77200187265877, -2.77200187265877. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-3%2Ax%2B-16+%29



I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J