Question 560575


{{{x^2+8x-20=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+8x-20}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=8}}}, and {{{C=-20}}}



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



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



{{{x = (-(8) +- sqrt( (8)^2-4(1)(-20) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=8}}}, and {{{C=-20}}}



{{{x = (-8 +- sqrt( 64-4(1)(-20) ))/(2(1))}}} Square {{{8}}} to get {{{64}}}. 



{{{x = (-8 +- sqrt( 64--80 ))/(2(1))}}} Multiply {{{4(1)(-20)}}} to get {{{-80}}}



{{{x = (-8 +- sqrt( 64+80 ))/(2(1))}}} Rewrite {{{sqrt(64--80)}}} as {{{sqrt(64+80)}}}



{{{x = (-8 +- sqrt( 144 ))/(2(1))}}} Add {{{64}}} to {{{80}}} to get {{{144}}}



{{{x = (-8 +- sqrt( 144 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-8 +- 12)/(2)}}} Take the square root of {{{144}}} to get {{{12}}}. 



{{{x = (-8 + 12)/(2)}}} or {{{x = (-8 - 12)/(2)}}} Break up the expression. 



{{{x = (4)/(2)}}} or {{{x =  (-20)/(2)}}} Combine like terms. 



{{{x = 2}}} or {{{x = -10}}} Simplify. 



So the solutions are {{{x = 2}}} or {{{x = -10}}} 

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