Question 600662


{{{3x^2-2x-7=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2-2x-7}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-2}}}, and {{{C=-7}}}



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



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



{{{x = (-(-2) +- sqrt( (-2)^2-4(3)(-7) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-2}}}, and {{{C=-7}}}



{{{x = (2 +- sqrt( (-2)^2-4(3)(-7) ))/(2(3))}}} Negate {{{-2}}} to get {{{2}}}. 



{{{x = (2 +- sqrt( 4-4(3)(-7) ))/(2(3))}}} Square {{{-2}}} to get {{{4}}}. 



{{{x = (2 +- sqrt( 4--84 ))/(2(3))}}} Multiply {{{4(3)(-7)}}} to get {{{-84}}}



{{{x = (2 +- sqrt( 4+84 ))/(2(3))}}} Rewrite {{{sqrt(4--84)}}} as {{{sqrt(4+84)}}}



{{{x = (2 +- sqrt( 88 ))/(2(3))}}} Add {{{4}}} to {{{84}}} to get {{{88}}}



{{{x = (2 +- sqrt( 88 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (2 +- 2*sqrt(22))/(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 = (2+2*sqrt(22))/(6)}}} or {{{x = (2-2*sqrt(22))/(6)}}} Break up the expression.  



{{{x = (1+sqrt(22))/(3)}}} or {{{x = (1-sqrt(22))/(3)}}} Reduce



So the solutions are {{{x = (1+sqrt(22))/(3)}}} or {{{x = (1-sqrt(22))/(3)}}}