Question 120811
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+7*x+12=0}}} ( notice {{{a=1}}}, {{{b=7}}}, and {{{c=12}}})





{{{x = (-7 +- sqrt( (7)^2-4*1*12 ))/(2*1)}}} Plug in a=1, b=7, and c=12




{{{x = (-7 +- sqrt( 49-4*1*12 ))/(2*1)}}} Square 7 to get 49  




{{{x = (-7 +- sqrt( 49+-48 ))/(2*1)}}} Multiply {{{-4*12*1}}} to get {{{-48}}}




{{{x = (-7 +- sqrt( 1 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




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


So now the expression breaks down into two parts


{{{x = (-7 + 1)/2}}} or {{{x = (-7 - 1)/2}}}


Lets look at the first part:


{{{x=(-7 + 1)/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=(-7 - 1)/2}}}


{{{x=-8/2}}} Subtract the terms in the numerator

{{{x=-4}}} Divide


So another answer is

{{{x=-4}}}


So our solutions are:

{{{x=-3}}} or {{{x=-4}}}


which look like this in {{{a+bi}}} form


{{{x=-3+0i}}} or {{{x=-4+0i}}}





Notice when we graph {{{x^2+7*x+12}}}, we get:


{{{ graph( 500, 500, -14, 7, -14, 7,1*x^2+7*x+12) }}}


and we can see that the roots are {{{x=-3}}} and {{{x=-4}}}. This verifies our answer