Question 266123


{{{2r^2+10r+11=0}}} Start with the given equation.



Notice that the quadratic {{{2r^2+10r+11}}} is in the form of {{{Ar^2+Br+C}}} where {{{A=2}}}, {{{B=10}}}, and {{{C=11}}}



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



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



{{{r = (-(10) +- sqrt( (10)^2-4(2)(11) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=10}}}, and {{{C=11}}}



{{{r = (-10 +- sqrt( 100-4(2)(11) ))/(2(2))}}} Square {{{10}}} to get {{{100}}}. 



{{{r = (-10 +- sqrt( 100-88 ))/(2(2))}}} Multiply {{{4(2)(11)}}} to get {{{88}}}



{{{r = (-10 +- sqrt( 12 ))/(2(2))}}} Subtract {{{88}}} from {{{100}}} to get {{{12}}}



{{{r = (-10 +- sqrt( 12 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{r = (-10 +- 2*sqrt(3))/(4)}}} 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>)  



{{{r = (-10+2*sqrt(3))/(4)}}} or {{{r = (-10-2*sqrt(3))/(4)}}} Break up the expression.  



So the solutions are {{{r = (-10+2*sqrt(3))/(4)}}} or {{{r = (-10-2*sqrt(3))/(4)}}}