SOLUTION: Do I leave my answer in fraction form or in whole number form? How do I know? x^2-3x-28=0 -3+-sqrt((3)^2-4(1)(-28))/(2(1)) -3+-sqrt(9+112)/(2) -3+-sqrt(121)/(2) x=(4)/(1)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Do I leave my answer in fraction form or in whole number form? How do I know? x^2-3x-28=0 -3+-sqrt((3)^2-4(1)(-28))/(2(1)) -3+-sqrt(9+112)/(2) -3+-sqrt(121)/(2) x=(4)/(1)      Log On


   

Question 411281: Do I leave my answer in fraction form or in whole number form? How do I know?
x^2-3x-28=0
-3+-sqrt((3)^2-4(1)(-28))/(2(1))
-3+-sqrt(9+112)/(2)
-3+-sqrt(121)/(2)
x=(4)/(1) and x=-(7)/(1)
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Do I leave my answer in fraction form or in whole number form? How do I know?
x^2-3x-28=0
-3+-sqrt((3)^2-4(1)(-28))/(2(1))
-3+-sqrt(9+112)/(2)
-3+-sqrt(121)/(2)
x=(4)/(1) and x=-(7)/(1)
--------------------------------
Should be:
x = [-b +- sqrt(b^2-4ac)]/(2a)
---
x = [3 +- sqrt(9-4*1*-28)]/(2
---
x = [3 +- sqrt(121)]/2
---
x = [3+-11]/2
----
x = -8/2 or x = 14/2
---
x = -4 or x = 7
====================
Cheers,
Stan H.
===================

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


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




So lets solve x%5E2-3%2Ax-28=0 ( notice a=1, b=-3, and c=-28)





x+=+%28--3+%2B-+sqrt%28+%28-3%29%5E2-4%2A1%2A-28+%29%29%2F%282%2A1%29 Plug in a=1, b=-3, and c=-28




x+=+%283+%2B-+sqrt%28+%28-3%29%5E2-4%2A1%2A-28+%29%29%2F%282%2A1%29 Negate -3 to get 3




x+=+%283+%2B-+sqrt%28+9-4%2A1%2A-28+%29%29%2F%282%2A1%29 Square -3 to get 9 (note: remember when you square -3, you must square the negative as well. This is because %28-3%29%5E2=-3%2A-3=9.)




x+=+%283+%2B-+sqrt%28+9%2B112+%29%29%2F%282%2A1%29 Multiply -4%2A-28%2A1 to get 112




x+=+%283+%2B-+sqrt%28+121+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%283+%2B-+11%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%283+%2B-+11%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


x+=+%283+%2B+11%29%2F2 or x+=+%283+-+11%29%2F2


Lets look at the first part:


x=%283+%2B+11%29%2F2


x=14%2F2 Add the terms in the numerator

x=7 Divide


So one answer is

x=7




Now lets look at the second part:


x=%283+-+11%29%2F2


x=-8%2F2 Subtract the terms in the numerator

x=-4 Divide


So another answer is

x=-4


So our solutions are:

x=7 or x=-4