Question 174722


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



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



{{{x = (-4 +- sqrt( 16--36 ))/(2(3))}}} Multiply {{{4(3)(-3)}}} to get {{{-36}}}



{{{x = (-4 +- sqrt( 16+36 ))/(2(3))}}} Rewrite {{{sqrt(16--36)}}} as {{{sqrt(16+36)}}}



{{{x = (-4 +- sqrt( 52 ))/(2(3))}}} Add {{{16}}} to {{{36}}} to get {{{52}}}



{{{x = (-4 +- sqrt( 52 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-4 +- 2*sqrt(13))/(6)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-4+2*sqrt(13))/(6)}}} or {{{x = (-4-2*sqrt(13))/(6)}}} Break up the expression. 




{{{x = (-2+sqrt(13))/(3)}}} or {{{x = (-2-sqrt(13))/(3)}}} Reduce.



So the answers are {{{x = (-2+sqrt(13))/(3)}}} or {{{x = (-2-sqrt(13))/(3)}}}  



which approximate to {{{x=0.535}}} or {{{x=-1.869}}}