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
x= 3-27 = -24 = -12 or x= 3+2
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Question 295727: 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= 3-27 = -24 = -12 or x= 3+27 = 30 = 15
all divide by 2
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The first thing you did that I wouldn't have done is choose to solve this quadratic using the quadratic formula. Not that it is actually wrong to do so, it is just that factoring this one is so much easier. However, since you chose to use the quadratic formula, let's do it properly.
\ \pm\ \sqrt{(-3)^2\ -\ 4(1)(-28)}}{2(1)} )
}}{2} )
or 
Now, had you simply recognized that 
and 
, you could have written:
(x\ +\ 4)\ =\ 0)
and therefore 
or 
John
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