Question 407298
{{{r^2+3r=8}}} Start with the given equation.



{{{r^2+3r-8=0}}} Subtract 8 from both sides.



Notice we have a quadratic equation in the form of {{{ar^2+br+c}}} where {{{a=1}}}, {{{b=3}}}, and {{{c=-8}}}



Let's use the quadratic formula to solve for r



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



{{{r = (-(3) +- sqrt( (3)^2-4(1)(-8) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=3}}}, and {{{c=-8}}}



{{{r = (-3 +- sqrt( 9-4(1)(-8) ))/(2(1))}}} Square {{{3}}} to get {{{9}}}. 



{{{r = (-3 +- sqrt( 9--32 ))/(2(1))}}} Multiply {{{4(1)(-8)}}} to get {{{-32}}}



{{{r = (-3 +- sqrt( 9+32 ))/(2(1))}}} Rewrite {{{sqrt(9--32)}}} as {{{sqrt(9+32)}}}



{{{r = (-3 +- sqrt( 41 ))/(2(1))}}} Add {{{9}}} to {{{32}}} to get {{{41}}}



{{{r = (-3 +- sqrt( 41 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{r = (-3+sqrt(41))/(2)}}} or {{{r = (-3-sqrt(41))/(2)}}} Break up the expression.  



So the answers are {{{r = (-3+sqrt(41))/(2)}}} or {{{r = (-3-sqrt(41))/(2)}}} 



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