Question 143165
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+6*x-8=0}}} ( notice {{{a=1}}}, {{{b=6}}}, and {{{c=-8}}})





{{{x = (-6 +- sqrt( (6)^2-4*1*-8 ))/(2*1)}}} Plug in a=1, b=6, and c=-8




{{{x = (-6 +- sqrt( 36-4*1*-8 ))/(2*1)}}} Square 6 to get 36  




{{{x = (-6 +- sqrt( 36+32 ))/(2*1)}}} Multiply {{{-4*-8*1}}} to get {{{32}}}




{{{x = (-6 +- sqrt( 68 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-6 +- 2*sqrt(17))/(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 = (-6 +- 2*sqrt(17))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-6 + 2*sqrt(17))/2}}} or {{{x = (-6 - 2*sqrt(17))/2}}}



Now break up the fraction



{{{x=-6/2+2*sqrt(17)/2}}} or {{{x=-6/2-2*sqrt(17)/2}}}



Simplify



{{{x=-3+sqrt(17)}}} or {{{x=-3-sqrt(17)}}}




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Answer:


So the solutions are


{{{x=-3+sqrt(17)}}} or {{{x=-3-sqrt(17)}}}