Question 210914


{{{3x^2=8x+9}}} Start with the given equation.



{{{3x^2-8x-9=0}}} Get every term to the left side.



Notice that the quadratic {{{3x^2-8x-9}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-8}}}, and {{{C=-9}}}



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)(-9) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-8}}}, and {{{C=-9}}}



{{{x = (8 +- sqrt( (-8)^2-4(3)(-9) ))/(2(3))}}} Negate {{{-8}}} to get {{{8}}}. 



{{{x = (8 +- sqrt( 64-4(3)(-9) ))/(2(3))}}} Square {{{-8}}} to get {{{64}}}. 



{{{x = (8 +- sqrt( 64--108 ))/(2(3))}}} Multiply {{{4(3)(-9)}}} to get {{{-108}}}



{{{x = (8 +- sqrt( 64+108 ))/(2(3))}}} Rewrite {{{sqrt(64--108)}}} as {{{sqrt(64+108)}}}



{{{x = (8 +- sqrt( 172 ))/(2(3))}}} Add {{{64}}} to {{{108}}} to get {{{172}}}



{{{x = (8 +- sqrt( 172 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (8 +- 2*sqrt(43))/(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 = (8+2*sqrt(43))/(6)}}} or {{{x = (8-2*sqrt(43))/(6)}}} Break up the expression.  



{{{x = (4+sqrt(43))/(3)}}} or {{{x = (4-sqrt(43))/(3)}}} Reduce



So the solutions are {{{x = (4+sqrt(43))/(3)}}} or {{{x = (4-sqrt(43))/(3)}}}