Question 632702
{{{3x^2+19x-14=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2+19x-14}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=19}}}, and {{{C=-14}}}



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



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



{{{x = (-(19) +- sqrt( (19)^2-4(3)(-14) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=19}}}, and {{{C=-14}}}



{{{x = (-19 +- sqrt( 361-4(3)(-14) ))/(2(3))}}} Square {{{19}}} to get {{{361}}}. 



{{{x = (-19 +- sqrt( 361--168 ))/(2(3))}}} Multiply {{{4(3)(-14)}}} to get {{{-168}}}



{{{x = (-19 +- sqrt( 361+168 ))/(2(3))}}} Rewrite {{{sqrt(361--168)}}} as {{{sqrt(361+168)}}}



{{{x = (-19 +- sqrt( 529 ))/(2(3))}}} Add {{{361}}} to {{{168}}} to get {{{529}}}



{{{x = (-19 +- sqrt( 529 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-19 +- 23)/(6)}}} Take the square root of {{{529}}} to get {{{23}}}. 



{{{x = (-19 + 23)/(6)}}} or {{{x = (-19 - 23)/(6)}}} Break up the expression. 



{{{x = (4)/(6)}}} or {{{x =  (-42)/(6)}}} Combine like terms. 



{{{x = 2/3}}} or {{{x = -7}}} Simplify. 



So the solutions are {{{x = 2/3}}} or {{{x = -7}}} 


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