Question 150810
From {{{3x^2+19x-14}}}, we can see that {{{a=3}}}, {{{b=19}}}, and {{{c=-14}}}



{{{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 our answers are {{{x = 2/3}}} or {{{x = -7}}}