Question 472158


{{{3x^2+22x-16=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2+22x-16}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=22}}}, and {{{C=-16}}}



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



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



{{{x = (-(22) +- sqrt( (22)^2-4(3)(-16) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=22}}}, and {{{C=-16}}}



{{{x = (-22 +- sqrt( 484-4(3)(-16) ))/(2(3))}}} Square {{{22}}} to get {{{484}}}. 



{{{x = (-22 +- sqrt( 484--192 ))/(2(3))}}} Multiply {{{4(3)(-16)}}} to get {{{-192}}}



{{{x = (-22 +- sqrt( 484+192 ))/(2(3))}}} Rewrite {{{sqrt(484--192)}}} as {{{sqrt(484+192)}}}



{{{x = (-22 +- sqrt( 676 ))/(2(3))}}} Add {{{484}}} to {{{192}}} to get {{{676}}}



{{{x = (-22 +- sqrt( 676 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-22 +- 26)/(6)}}} Take the square root of {{{676}}} to get {{{26}}}. 



{{{x = (-22 + 26)/(6)}}} or {{{x = (-22 - 26)/(6)}}} Break up the expression. 



{{{x = (4)/(6)}}} or {{{x =  (-48)/(6)}}} Combine like terms. 



{{{x = 2/3}}} or {{{x = -8}}} Simplify. 



So the solutions are {{{x = 2/3}}} or {{{x = -8}}}