Question 151294


{{{3x^2+8x-3=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=3}}}, {{{b=8}}}, and {{{c=-3}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(8) +- sqrt( (8)^2-4(3)(-3) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=8}}}, and {{{c=-3}}}



{{{x = (-8 +- sqrt( 64-4(3)(-3) ))/(2(3))}}} Square {{{8}}} to get {{{64}}}. 



{{{x = (-8 +- sqrt( 64--36 ))/(2(3))}}} Multiply {{{4(3)(-3)}}} to get {{{-36}}}



{{{x = (-8 +- sqrt( 64+36 ))/(2(3))}}} Rewrite {{{sqrt(64--36)}}} as {{{sqrt(64+36)}}}



{{{x = (-8 +- sqrt( 100 ))/(2(3))}}} Add {{{64}}} to {{{36}}} to get {{{100}}}



{{{x = (-8 +- sqrt( 100 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-8 +- 10)/(6)}}} Take the square root of {{{100}}} to get {{{10}}}. 



{{{x = (-8 + 10)/(6)}}} or {{{x = (-8 - 10)/(6)}}} Break up the expression. 



{{{x = (2)/(6)}}} or {{{x =  (-18)/(6)}}} Combine like terms. 



{{{x = 1/3}}} or {{{x = -3}}} Simplify. 



So our answers are {{{x = 1/3}}} or {{{x = -3}}}