Question 76944
*[invoke quadratic_factoring 5, 17, 6]

So our quadratic factors to:

{{{(5x+2)(x+3)=0}}}

Now set each factor equal to zero

{{{5x+2=0}}}
{{{5x=-2}}}
{{{x=-2/5}}}
{{{x+3=0}}}
{{{x=-3}}}
So our answers are 
{{{x=-2/5}}} or {{{x=-3}}}



or you could use the quadratic formula to solve for x:

*[invoke quadratic "x", 5, 17, 6 ]