Question 461844


{{{2x^2+32x-34=0}}} Start with the given equation.



Notice that the quadratic {{{2x^2+32x-34}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=32}}}, and {{{C=-34}}}



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



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



{{{x = (-(32) +- sqrt( (32)^2-4(2)(-34) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=32}}}, and {{{C=-34}}}



{{{x = (-32 +- sqrt( 1024-4(2)(-34) ))/(2(2))}}} Square {{{32}}} to get {{{1024}}}. 



{{{x = (-32 +- sqrt( 1024--272 ))/(2(2))}}} Multiply {{{4(2)(-34)}}} to get {{{-272}}}



{{{x = (-32 +- sqrt( 1024+272 ))/(2(2))}}} Rewrite {{{sqrt(1024--272)}}} as {{{sqrt(1024+272)}}}



{{{x = (-32 +- sqrt( 1296 ))/(2(2))}}} Add {{{1024}}} to {{{272}}} to get {{{1296}}}



{{{x = (-32 +- sqrt( 1296 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-32 +- 36)/(4)}}} Take the square root of {{{1296}}} to get {{{36}}}. 



{{{x = (-32 + 36)/(4)}}} or {{{x = (-32 - 36)/(4)}}} Break up the expression. 



{{{x = (4)/(4)}}} or {{{x =  (-68)/(4)}}} Combine like terms. 



{{{x = 1}}} or {{{x = -17}}} Simplify. 



So the solutions are {{{x = 1}}} or {{{x = -17}}}