Question 202110

{{{c^2-4=0}}} Start with the given equation.



Notice that the quadratic {{{c^2-4}}} is in the form of {{{Ac^2+Bc+C}}} where {{{A=1}}}, {{{B=0}}}, and {{{C=-4}}}



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



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



{{{c = (0 +- sqrt( (0)^2-4(1)(-4) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=0}}}, and {{{C=-4}}}



{{{c = (0 +- sqrt( 0-4(1)(-4) ))/(2(1))}}} Square {{{0}}} to get {{{0}}}. 



{{{c = (0 +- sqrt( 0--16 ))/(2(1))}}} Multiply {{{4(1)(-4)}}} to get {{{-16}}}



{{{c = (0 +- sqrt( 0+16 ))/(2(1))}}} Rewrite {{{sqrt(0--16)}}} as {{{sqrt(0+16)}}}



{{{c = (0 +- sqrt( 16 ))/(2(1))}}} Add {{{0}}} to {{{16}}} to get {{{16}}}



{{{c = (0 +- sqrt( 16 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{c = (0 +- 4)/(2)}}} Take the square root of {{{16}}} to get {{{4}}}. 



{{{c = (0 + 4)/(2)}}} or {{{c = (0 - 4)/(2)}}} Break up the expression. 



{{{c = (4)/(2)}}} or {{{c =  (-4)/(2)}}} Combine like terms. 



{{{c = 2}}} or {{{c = -2}}} Simplify. 



So the solutions are {{{c = 2}}} or {{{c = -2}}}