Question 195548
{{{8c^2+21c+10=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ac^2+bc+c}}} where {{{a=8}}}, {{{b=21}}}, and {{{c=10}}}



Let's use the quadratic formula to solve for c



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



{{{c = (-(21) +- sqrt( (21)^2-4(8)(10) ))/(2(8))}}} Plug in  {{{a=8}}}, {{{b=21}}}, and {{{c=10}}}



{{{c = (-21 +- sqrt( 441-4(8)(10) ))/(2(8))}}} Square {{{21}}} to get {{{441}}}. 



{{{c = (-21 +- sqrt( 441-320 ))/(2(8))}}} Multiply {{{4(8)(10)}}} to get {{{320}}}



{{{c = (-21 +- sqrt( 121 ))/(2(8))}}} Subtract {{{320}}} from {{{441}}} to get {{{121}}}



{{{c = (-21 +- sqrt( 121 ))/(16)}}} Multiply {{{2}}} and {{{8}}} to get {{{16}}}. 



{{{c = (-21 +- 11)/(16)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{c = (-21 + 11)/(16)}}} or {{{c = (-21 - 11)/(16)}}} Break up the expression. 



{{{c = (-10)/(16)}}} or {{{c =  (-32)/(16)}}} Combine like terms. 



{{{c = -5/8}}} or {{{c = -2}}} Simplify. 



So the answers are {{{c = -5/8}}} or {{{c = -2}}}