Question 170747
<pre>
I need some help with this problem.
i asked my friend to help me with a previous problem and he ended up just giving me the answer cause i still don't get it.
The one already answered == 
"":How i tried to work it out.
[]:my answer

1. {{{ 2x^2-10x+12=0 }}}
"2(x+6)(x-1)"
[x=6,(-1)]
Was that correct?

2. {{{ x^2+2x=-11 }}}
?????

Please help me.
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For 1. {{{2x^2 - 10x + 12 = 0}}}, your factorization to get: "2(x+6)(x-1)" is WRONG!! Your solutions: [x=6,(-1)] are also WRONG! 
FOILing 2(x + 6)(x - 1), we get: {{{2(x^2 + 5x - 6)}}} = {{{2x^2 + 10x - 12}}}, which doesn't correspond to the given trinomial, {{{2x^2 - 10x + 12}}}

The person who responded also has a different factorized form [(x-6)(x+1)=0--], but unfortunately, this is also WRONG! FOILing
(x - 6)(x + 1) = {{{x^2 - 5x - 6}}}, which doesn't correspond with the reduced trinomial, {{{x^2 - 5x + 6}}} (see below). His/her solutions,
x = 6, and x = - 1 are obviously WRONG too!

The correct factorization and solutions follow!
{{{2x^2 - 10x + 12 = 0}}}
{{{2(x^2 - 5x + 6) = 0}}}
{{{x^2 - 5x + 6 = 0}}}
(x - 3)(x - 2) = 0 <==== Correct factorized form
x - 3 = 0       OR       x - 2 = 0
    x = 3       OR           x = 2 <===== CORRECT solutions
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Problem 2. {{{x^2 + 2x = -11}}} CANNOT be factorized using integers. So, to get its roots/solutions, you'll need to use
the QUADRATIC EQUATION formula</pre>