Question 98883
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+9*x+18=0}}} ( notice {{{a=1}}}, {{{b=9}}}, and {{{c=18}}})


{{{x = (-9 +- sqrt( (9)^2-4*1*18 ))/(2*1)}}} Plug in a=1, b=9, and c=18




{{{x = (-9 +- sqrt( 81-4*1*18 ))/(2*1)}}} Square 9 to get 81  




{{{x = (-9 +- sqrt( 81+-72 ))/(2*1)}}} Multiply {{{-4*18*1}}} to get {{{-72}}}




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




{{{x = (-9 +- 3)/(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 = (-9 +- 3)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-9 + 3)/2}}} or {{{x = (-9 - 3)/2}}}


Lets look at the first part:


{{{x=(-9 + 3)/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=(-9 - 3)/2}}}


{{{x=-12/2}}} Subtract the terms in the numerator

{{{x=-6}}} Divide


So another answer is

{{{x=-6}}}


So our solutions are:

{{{x=-3}}} or {{{x=-6}}}


Notice when we graph {{{x^2+9*x+18}}}, we get:


{{{ graph( 500, 500, -16, 7, -16, 7,1*x^2+9*x+18) }}}


and we can see that the roots are {{{x=-3}}} and {{{x=-6}}}. This verifies our answer