Question 76350
a)
*[invoke factoring_quadratics 1, -6, 8]
So in other words, {{{x^2-6x+8=0}}} factors to {{{(x-2)(x-4)=0}}}
Now set each factor equal to zero:
{{{x-2=0}}}
{{{x=2}}} Solve for x
{{{x-4=0}}}
{{{x=4}}} Solve for x
So our solution is x=2 and x=4





b)
*[invoke quadratic "x", 1, -6, 8 ]
So our solution is x=2, and x=4. 
Notice how we got the same answer but took another route to get there.