Question 88859
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^2-5*x-6=0}}} ( notice {{{a=1}}}, {{{b=-5}}}, and {{{c=-6}}})


{{{x = (--5 +- sqrt( (-5)^2-4*1*-6 ))/(2*1)}}} Plug in a=1, b=-5, and c=-6




{{{x = (5 +- sqrt( (-5)^2-4*1*-6 ))/(2*1)}}} Negate -5 to get 5




{{{x = (5 +- sqrt( 25-4*1*-6 ))/(2*1)}}} Square -5 to get 25




{{{x = (5 +- sqrt( 25+24 ))/(2*1)}}} Multiply {{{-4*-6*1}}} to get {{{24}}}




{{{x = (5 +- sqrt( 49 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (5 +- 7)/(2*1)}}} Simplify the square root




{{{x = (5 +- 7)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (5 + 7)/2}}} or {{{x = (5 - 7)/2}}}


Lets look at the first part:


{{{x=(5 + 7)/2}}}


{{{x=12/2}}} Add the terms in the numerator

{{{x=6}}} Divide


So one answer is

{{{x=6}}}




Now lets look at the second part:


{{{x=(5 - 7)/2}}}


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

{{{x=-1}}} Divide


So another answer is

{{{x=-1}}}


So our solutions are:

{{{x=6}}} or {{{x=-1}}}


Notice when we graph {{{x^2-5*x-6}}}, we get:


{{{ graph( 500, 500, -11, 16, -11, 16,1*x^2+-5*x+-6) }}}


and we can see that the roots are {{{x=6}}} and {{{x=-1}}}. This verifies our answer