Question 181059
{{{6x^2 + 14x + 4}}}
To locate the roots, set the equation equal to {{{0}}}
{{{6x^2 + 14x + 4 = 0}}}
Divide both sides by {{{2}}}
{{{3x^2 + 7x + 2 = 0}}}
I often try to factor by trying combinations that will
factor the outside terms and try to get the inside term,
for instance:
(3x + ?)(x + ?) I've got the 1st term factored, now for the 3rd
(3x + 2)(x + 1) Does this work?
{{{(3x + 2)(x + 1) = 3x^2 + 5x + 2}}} not quite
How about
(3x + 1)(x + 2) ?
{{{(3x + 1)(x + 2) = 3x^2 + 7x + 2}}} that's it
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Some are not this neat, and you have to use the quadratic equation
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
Use this when the form of the equation is
{{{ax^2 + bx + c = 0}}}
{{{a = 3}}}
{{{b = 7}}}
{{{c = 2}}}
Some are not this neat, and you have to use the quadratic equation
{{{x = (-7 +- sqrt( 7^2-4*3*2 ))/(2*3) }}} 
{{{x = (-7 +- sqrt( 49 - 24 )) / 6 }}}
{{{x = (-7 +- sqrt(25) ) / 6 }}} 
{{{x = (-7 +- 5 ) / 6 }}}
{{{x = -(7/6) + (5/6)}}}
{{{x = -(1/3)}}}
{{{(x + (1/3)) = 0}}}
and
{{{x = -(7/6) - (5/6)}}}
{{{x = -(12/6)}}}
{{{x = -2}}}
{{{(x + 2) = 0}}}
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{{{(x + (1/3))(x + 2) = 0}}}
{{{x^2 + (7/3)x + (2/3) = 0}}}
Multiply both sides by {{{3}}}
{{{3x^2 + 7x + 2 = 0}}}