SOLUTION: This is how I solved the problem. Where did I mis-calculate? {{{x^2-3x-28=0}}} -3(-3)+- sq.rt(-3)^2-4`1`(-28) 3 +- sq.rt.729 3+9 ___________ ___________ _____

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Question 295725: This is how I solved the problem. Where did I mis-calculate?

-3(-3)+- sq.rt(-3)^2-4`1`(-28) 3 +- sq.rt.729 3+9
______________________________ = x= ______________ = _____
2`1 2 2
x= 3-27 = -24 = -12 or x= 3+27 = 30 = 15
____ ___ ____ __ -12,15
2 2 2 2

Answer by jim_thompson5910(35256) (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

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=1, b=-3, and c=-28

Negate -3 to get 3

Square -3 to get 9 (note: remember when you square -3, you must square the negative as well. This is because .)

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

or

Lets look at the first part:

Add the terms in the numerator

Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator

Divide

So another answer is

So our solutions are:

or

SOLUTION: This is how I solved the problem. Where did I mis-calculate? {{{x^2-3x-28=0}}} -3(-3)+- sq.rt(-3)^2-4`1`(-28) 3 +- sq.rt.729 3+9 ___________ ___________ ______ =

SOLUTION: This is how I solved the problem. Where did I mis-calculate? {{{x^2-3x-28=0}}} -3(-3)+- sq.rt(-3)^2-4`1`(-28) 3 +- sq.rt.729 3+9 _____ _ __ =