Question 150922


{{{x^2-2x-5=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-2}}}, and {{{c=-5}}}



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)(-5) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-2}}}, and {{{c=-5}}}



{{{x = (2 +- sqrt( (-2)^2-4(1)(-5) ))/(2(1))}}} Negate {{{-2}}} to get {{{2}}}. 



{{{x = (2 +- sqrt( 4-4(1)(-5) ))/(2(1))}}} Square {{{-2}}} to get {{{4}}}. 



{{{x = (2 +- sqrt( 4--20 ))/(2(1))}}} Multiply {{{4(1)(-5)}}} to get {{{-20}}}



{{{x = (2 +- sqrt( 4+20 ))/(2(1))}}} Rewrite {{{sqrt(4--20)}}} as {{{sqrt(4+20)}}}



{{{x = (2 +- sqrt( 24 ))/(2(1))}}} Add {{{4}}} to {{{20}}} to get {{{24}}}



{{{x = (2 +- sqrt( 24 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (2 +- 2*sqrt(6))/(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(6))/(2)}}} Break up the fraction.  



{{{x = 1 +- sqrt(6)}}} Reduce.  



{{{x = 1+sqrt(6)}}} or {{{x = 1-sqrt(6)}}} Break up the expression.  



So our answers are {{{x = 1+sqrt(6)}}} or {{{x = 1-sqrt(6)}}} 



which approximate to {{{x=3.449}}} or {{{x=-1.449}}}