Question 687620
{{{ 6x^2+13x-5=0 }}}
Use the quadratic formula when the equation
is in the form {{{ a*x^2 + b*x + c = 0 }}}
{{{ x = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 6 }}}
{{{ b = 13 }}}
{{{ c = -5 }}}
{{{ x = (-13 +- sqrt( 13^2 - 4*6*(-5) )) / (2*6) }}}
{{{ x = (-13 +- sqrt( 169 +120 )) / 12 }}}
{{{ x = (-13 +- sqrt( 289 )) / 12 }}}
{{{ x = (-13 +- 17 ) / 12 }}}
{{{ x = ( -13 + 17 ) / 12 }}}
{{{ x = 4/12 }}}
(1) {{{ x = 1/3 }}}
and, also,
{{{ x = ( -13 - 17 ) / 12 }}}
{{{ x = -30/12 }}}
(2) {{{ x = -5/2 }}}
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Now rewrite (1) and (2) as:
(1) {{{ x - 1/3 = 0 }}}
(2) {{{ x + 5/2 = 0 }}}
Now multiply both sides of (1) by {{{ 3 }}} 
and both sides of (2) by {{{ 2 }}}
(1) {{{ 3x - 1 = 0 }}}
(2) {{{ 2x + 5 = 0 }}}
So, the factoring is:
{{{ ( 3x - 1 )*( 2x + 5 ) = 0 }}}
check:
{{{ 6x^2 - 2x + 15x - 5 = 0 }}}
{{{ 6^2 + 13x - 5 = 0 }}}
OK