Question 299444


{{{x^2-10x-22=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-10x-22}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-10}}}, and {{{C=-22}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-10) +- sqrt( (-10)^2-4(1)(-22) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-10}}}, and {{{C=-22}}}



{{{x = (10 +- sqrt( (-10)^2-4(1)(-22) ))/(2(1))}}} Negate {{{-10}}} to get {{{10}}}. 



{{{x = (10 +- sqrt( 100-4(1)(-22) ))/(2(1))}}} Square {{{-10}}} to get {{{100}}}. 



{{{x = (10 +- sqrt( 100--88 ))/(2(1))}}} Multiply {{{4(1)(-22)}}} to get {{{-88}}}



{{{x = (10 +- sqrt( 100+88 ))/(2(1))}}} Rewrite {{{sqrt(100--88)}}} as {{{sqrt(100+88)}}}



{{{x = (10 +- sqrt( 188 ))/(2(1))}}} Add {{{100}}} to {{{88}}} to get {{{188}}}



{{{x = (10 +- sqrt( 188 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (10 +- 2*sqrt(47))/(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 = (10)/(2) +- (2*sqrt(47))/(2)}}} Break up the fraction.  



{{{x = 5 +- sqrt(47)}}} Reduce.  



{{{x = 5+sqrt(47)}}} or {{{x = 5-sqrt(47)}}} Break up the expression.  



So the solutions are {{{x = 5+sqrt(47)}}} or {{{x = 5-sqrt(47)}}}