Question 185542
{{{(1/2)x^2 - x -2 =0}}} Start with the given equation.



{{{x^2 - 2x -4 =0}}} Multiply EVERY term by the LCD 2 to clear the fractions.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-2}}}, and {{{c=-4}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



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



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



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



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



{{{x = (2 +- sqrt( 4+16 ))/(2(1))}}} Rewrite {{{sqrt(4--16)}}} as {{{sqrt(4+16)}}}



{{{x = (2 +- sqrt( 20 ))/(2(1))}}} Add {{{4}}} to {{{16}}} to get {{{20}}}



{{{x = (2 +- sqrt( 20 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (2 +- 2*sqrt(5))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (2)/(2) +- (2*sqrt(5))/(2)}}} Break up the fraction.  



{{{x = 1 +- sqrt(5)}}} Reduce. 




So the solutions are  {{{x = 1 +- sqrt(5)}}}



This really means that the solutions are {{{x = 1+sqrt(5)}}} or {{{x = 1-sqrt(5)}}} (since the symbol *[Tex \LARGE \pm] tells you that you have a "plus" or a "minus" in the place of the *[Tex \LARGE \pm])