Question 85029
*[invoke quadratic_factoring 1, 5, -14]


So we now have


{{{(x-2)(x+7)=0}}}


Set each factor equal to zero


{{{x-2=0}}} or {{{x+7=0}}}


Solve for x in each case


{{{x=2}}} or {{{x=-7}}}


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*[invoke quadratic_factoring 3, -22, 7]


So we now have


{{{(3x-1)(x-7)=0}}}


Set each factor equal to zero


{{{3x-1=0}}} or {{{x-7=0}}}


Solve for x in each case


{{{x=1/3}}} or {{{x=7}}}