Question 174541


{{{2(x+1)^2+14(x+1)+20=0}}} Start with the given equation.



Let {{{z=x+1}}}



{{{2z^2+14z+20=0}}} Replace each {{{x+1}}} with "z"



Notice we have a quadratic equation in the form of {{{az^2+bz+c}}} where {{{a=2}}}, {{{b=14}}}, and {{{c=20}}}



Let's use the quadratic formula to solve for z



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



{{{z = (-(14) +- sqrt( (14)^2-4(2)(20) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=14}}}, and {{{c=20}}}



{{{z = (-14 +- sqrt( 196-4(2)(20) ))/(2(2))}}} Square {{{14}}} to get {{{196}}}. 



{{{z = (-14 +- sqrt( 196-160 ))/(2(2))}}} Multiply {{{4(2)(20)}}} to get {{{160}}}



{{{z = (-14 +- sqrt( 36 ))/(2(2))}}} Subtract {{{160}}} from {{{196}}} to get {{{36}}}



{{{z = (-14 +- sqrt( 36 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{z = (-14 +- 6)/(4)}}} Take the square root of {{{36}}} to get {{{6}}}. 



{{{z = (-14 + 6)/(4)}}} or {{{z = (-14 - 6)/(4)}}} Break up the expression. 



{{{z = (-8)/(4)}}} or {{{z =  (-20)/(4)}}} Combine like terms. 



{{{z = -2}}} or {{{z = -5}}} Simplify.



{{{x+1 = -2}}} or {{{x+1 = -5}}} Plug in {{{z=x+1}}} for each equation.



{{{x = -2-1}}} or {{{x = -5-1}}} Subtract 1 from both sides (for each equation).



{{{x = -3}}} or {{{x = -6}}} Combine like terms (for each equation).



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Answer:



So the solutions are {{{x = -3}}} or {{{x = -6}}}



Note: both of these solutions are real.