Question 141179
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^2-4*x-6=0}}} ( notice {{{a=1}}}, {{{b=-4}}}, and {{{c=-6}}})





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




{{{x = (4 +- sqrt( (-4)^2-4*1*-6 ))/(2*1)}}} Negate -4 to get 4




{{{x = (4 +- sqrt( 16-4*1*-6 ))/(2*1)}}} Square -4 to get 16  (note: remember when you square -4, you must square the negative as well. This is because {{{(-4)^2=-4*-4=16}}}.)




{{{x = (4 +- sqrt( 16+24 ))/(2*1)}}} Multiply {{{-4*-6*1}}} to get {{{24}}}




{{{x = (4 +- sqrt( 40 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




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


So now the expression breaks down into two parts


{{{x = (4 + 2*sqrt(10))/2}}} or {{{x = (4 - 2*sqrt(10))/2}}}



Now break up the fraction



{{{x=+4/2+2*sqrt(10)/2}}} or {{{x=+4/2-2*sqrt(10)/2}}}



Reduce



{{{x=2+sqrt(10)}}} or {{{x=2-sqrt(10)}}}





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Answer:


So our solutions are



{{{x=2+sqrt(10)}}} or {{{x=2-sqrt(10)}}}