Question 96896
Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{x^2-x-6=0}}} ( notice {{{a=1}}}, {{{b=-1}}}, and {{{c=-6}}})


{{{x = (--1 +- sqrt( (-1)^2-4*1*-6 ))/(2*1)}}} Plug in a=1, b=-1, and c=-6




{{{x = (1 +- sqrt( (-1)^2-4*1*-6 ))/(2*1)}}} Negate -1 to get 1




{{{x = (1 +- sqrt( 1-4*1*-6 ))/(2*1)}}} Square -1 to get 1  (note: remember when you square -1, you must square the negative as well. This is because {{{(-1)^2=-1*-1=1}}}.)




{{{x = (1 +- sqrt( 1+24 ))/(2*1)}}} Multiply {{{-4*-6*1}}} to get {{{24}}}




{{{x = (1 +- sqrt( 25 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (1 +- 5)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (1 +- 5)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (1 + 5)/2}}} or {{{x = (1 - 5)/2}}}


Lets look at the first part:


{{{x=(1 + 5)/2}}}


{{{x=6/2}}} Add the terms in the numerator

{{{x=3}}} Divide


So one answer is

{{{x=3}}}




Now lets look at the second part:


{{{x=(1 - 5)/2}}}


{{{x=-4/2}}} Subtract the terms in the numerator

{{{x=-2}}} Divide


So another answer is

{{{x=-2}}}


So our solutions are:

{{{x=3}}} or {{{x=-2}}}


Notice when we graph {{{x^2-x-6}}}, we get:


{{{ graph( 500, 500, -12, 13, -12, 13,1*x^2+-1*x+-6) }}}


and we can see that the roots are {{{x=3}}} and {{{x=-2}}}. This verifies our answer