Question 88233
{{{x^2-9x-4=6}}}


{{{x^2-9x-10=0}}} Subtract 6 from both sides


Now let's use the quadratic formula to solve for x:



Starting with the general quadratic


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


the general form of the quadratic equation is:


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


So lets solve {{{x^2-9*x-10=0}}} (notice {{{a=1}}}, {{{b=-9}}}, and {{{c=-10}}})


{{{x = (--9 +- sqrt( (-9)^2-4*1*-10 ))/(2*1)}}} Plug in a=1, b=-9, and c=-10




{{{x = (9 +- sqrt( (-9)^2-4*1*-10 ))/(2*1)}}} Negate -9 to get 9




{{{x = (9 +- sqrt( 81-4*1*-10 ))/(2*1)}}} Square -9 to get 81




{{{x = (9 +- sqrt( 81+40 ))/(2*1)}}} Multiply {{{-4*-10*1}}} to get {{{40}}}




{{{x = (9 +- sqrt( 121 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (9 +- 11)/(2*1)}}} Simplify the square root




{{{x = (9 +- 11)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (9 + 11)/2}}} or {{{x = (9 - 11)/2}}}


Lets look at the first part:


{{{x=20/2}}} Add the terms in the numerator

{{{x=10}}} Divide


So one answer is

{{{x=10}}}

Now lets look at the second part:


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

{{{x=-1}}} Divide


So another answer is

{{{x=-1}}}


So our solutions are:

{{{x=10}}} or {{{x=-1}}}


Notice when we graph {{{x^2-9*x-10}}} we get:


{{{ graph( 500, 500, -11, 20, -11, 20,1*x^2+-9*x+-10) }}}


and we can see that the roots are {{{x=10}}} and {{{x=-1}}}. This verifies our answer