Question 76539
*[invoke quadratic_factoring 3, 7, -6]

So the quadratic {{{3x^2+7x-6}}} factors to {{{(3x-2)(x+3)}}}

In other words

{{{3x^2+7x-6=(3x-2)(x+3)=0}}}

{{{(3x-2)(x+3)=0}}}
Set each factor equal to zero
{{{3x-2=0}}}Solve for x
{{{3x=2}}}
{{{x=2/3}}}
{{{x+3=0}}}Solve for x
{{{x=-3}}}

So our solutions are:
{{{x=2/3}}} and {{{x=-3}}} which are two real roots

We can see this if we graph {{{y=3x^2+7x-6}}}

{{{ graph( 300, 200, -6, 5, -10, 10, 3x^2+7x-6) }}} graph of {{{y=3x^2+7x-6}}}