Question 299538


{{{2x^2+6x-56=0}}} Start with the given equation.



Notice that the quadratic {{{2x^2+6x-56}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=6}}}, and {{{C=-56}}}



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



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



{{{x = (-(6) +- sqrt( (6)^2-4(2)(-56) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=6}}}, and {{{C=-56}}}



{{{x = (-6 +- sqrt( 36-4(2)(-56) ))/(2(2))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36--448 ))/(2(2))}}} Multiply {{{4(2)(-56)}}} to get {{{-448}}}



{{{x = (-6 +- sqrt( 36+448 ))/(2(2))}}} Rewrite {{{sqrt(36--448)}}} as {{{sqrt(36+448)}}}



{{{x = (-6 +- sqrt( 484 ))/(2(2))}}} Add {{{36}}} to {{{448}}} to get {{{484}}}



{{{x = (-6 +- sqrt( 484 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-6 +- 22)/(4)}}} Take the square root of {{{484}}} to get {{{22}}}. 



{{{x = (-6 + 22)/(4)}}} or {{{x = (-6 - 22)/(4)}}} Break up the expression. 



{{{x = (16)/(4)}}} or {{{x =  (-28)/(4)}}} Combine like terms. 



{{{x = 4}}} or {{{x = -7}}} Simplify. 



So the solutions are {{{x = 4}}} or {{{x = -7}}}