Question 241430


{{{4r^2-13r+10=0}}} Start with the given equation.



Notice that the quadratic {{{4r^2-13r+10}}} is in the form of {{{Ar^2+Br+C}}} where {{{A=4}}}, {{{B=-13}}}, and {{{C=10}}}



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



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



{{{r = (-(-13) +- sqrt( (-13)^2-4(4)(10) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-13}}}, and {{{C=10}}}



{{{r = (13 +- sqrt( (-13)^2-4(4)(10) ))/(2(4))}}} Negate {{{-13}}} to get {{{13}}}. 



{{{r = (13 +- sqrt( 169-4(4)(10) ))/(2(4))}}} Square {{{-13}}} to get {{{169}}}. 



{{{r = (13 +- sqrt( 169-160 ))/(2(4))}}} Multiply {{{4(4)(10)}}} to get {{{160}}}



{{{r = (13 +- sqrt( 9 ))/(2(4))}}} Subtract {{{160}}} from {{{169}}} to get {{{9}}}



{{{r = (13 +- sqrt( 9 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{r = (13 +- 3)/(8)}}} Take the square root of {{{9}}} to get {{{3}}}. 



{{{r = (13 + 3)/(8)}}} or {{{r = (13 - 3)/(8)}}} Break up the expression. 



{{{r = (16)/(8)}}} or {{{r =  (10)/(8)}}} Combine like terms. 



{{{r = 2}}} or {{{r = 5/4}}} Simplify. 



So the solutions are {{{r = 2}}} or {{{r = 5/4}}}