Question 527063


{{{2x^2+18x+28=0}}} Start with the given equation.



Notice that the quadratic {{{2x^2+18x+28}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=18}}}, and {{{C=28}}}



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



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



{{{x = (-(18) +- sqrt( (18)^2-4(2)(28) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=18}}}, and {{{C=28}}}



{{{x = (-18 +- sqrt( 324-4(2)(28) ))/(2(2))}}} Square {{{18}}} to get {{{324}}}. 



{{{x = (-18 +- sqrt( 324-224 ))/(2(2))}}} Multiply {{{4(2)(28)}}} to get {{{224}}}



{{{x = (-18 +- sqrt( 100 ))/(2(2))}}} Subtract {{{224}}} from {{{324}}} to get {{{100}}}



{{{x = (-18 +- sqrt( 100 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-18 +- 10)/(4)}}} Take the square root of {{{100}}} to get {{{10}}}. 



{{{x = (-18 + 10)/(4)}}} or {{{x = (-18 - 10)/(4)}}} Break up the expression. 



{{{x = (-8)/(4)}}} or {{{x =  (-28)/(4)}}} Combine like terms. 



{{{x = -2}}} or {{{x = -7}}} Simplify. 



So the solutions are {{{x = -2}}} or {{{x = -7}}} 



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