Question 775290


{{{3x^2+13x-10=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2+13x-10}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=13}}}, and {{{C=-10}}}



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



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



{{{x = (-(13) +- sqrt( (13)^2-4(3)(-10) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=13}}}, and {{{C=-10}}}



{{{x = (-13 +- sqrt( 169-4(3)(-10) ))/(2(3))}}} Square {{{13}}} to get {{{169}}}. 



{{{x = (-13 +- sqrt( 169--120 ))/(2(3))}}} Multiply {{{4(3)(-10)}}} to get {{{-120}}}



{{{x = (-13 +- sqrt( 169+120 ))/(2(3))}}} Rewrite {{{sqrt(169--120)}}} as {{{sqrt(169+120)}}}



{{{x = (-13 +- sqrt( 289 ))/(2(3))}}} Add {{{169}}} to {{{120}}} to get {{{289}}}



{{{x = (-13 +- sqrt( 289 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-13 +- 17)/(6)}}} Take the square root of {{{289}}} to get {{{17}}}. 



{{{x = (-13 + 17)/(6)}}} or {{{x = (-13 - 17)/(6)}}} Break up the expression. 



{{{x = (4)/(6)}}} or {{{x =  (-30)/(6)}}} Combine like terms. 



{{{x = 2/3}}} or {{{x = -5}}} Simplify. 



So the solutions are {{{x = 2/3}}} or {{{x = -5}}}