Question 102533
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+2*x-8=0}}} ( notice {{{a=1}}}, {{{b=2}}}, and {{{c=-8}}})





{{{x = (-2 +- sqrt( (2)^2-4*1*-8 ))/(2*1)}}} Plug in a=1, b=2, and c=-8




{{{x = (-2 +- sqrt( 4-4*1*-8 ))/(2*1)}}} Square 2 to get 4  




{{{x = (-2 +- sqrt( 4+32 ))/(2*1)}}} Multiply {{{-4*-8*1}}} to get {{{32}}}




{{{x = (-2 +- sqrt( 36 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-2 +- 6)/(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 = (-2 +- 6)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-2 + 6)/2}}} or {{{x = (-2 - 6)/2}}}


Lets look at the first part:


{{{x=(-2 + 6)/2}}}


{{{x=4/2}}} Add the terms in the numerator

{{{x=2}}} Divide.....4,2....2 / 1


So one answer is

{{{x=2}}}




Now lets look at the second part:


{{{x=(-2 - 6)/2}}}


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

{{{x=-4}}} Divide


So another answer is

{{{x=-4}}}


So our solutions are:

{{{x=2}}} or {{{x=-4}}}


Notice when we graph {{{x^2+2*x-8}}}, we get:


{{{ graph( 500, 500, -14, 12, -14, 12,1*x^2+2*x+-8) }}}


and we can see that the roots are {{{x=2}}} and {{{x=-4}}}. This verifies our answer