Question 197972


{{{3x^2-4x-5=0}}} Start with the given equation.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-4}}}, and {{{C=-5}}}



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)(-5) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-4}}}, and {{{C=-5}}}



{{{x = (4 +- sqrt( (-4)^2-4(3)(-5) ))/(2(3))}}} Negate {{{-4}}} to get {{{4}}}. 



{{{x = (4 +- sqrt( 16-4(3)(-5) ))/(2(3))}}} Square {{{-4}}} to get {{{16}}}. 



{{{x = (4 +- sqrt( 16--60 ))/(2(3))}}} Multiply {{{4(3)(-5)}}} to get {{{-60}}}



{{{x = (4 +- sqrt( 16+60 ))/(2(3))}}} Rewrite {{{sqrt(16--60)}}} as {{{sqrt(16+60)}}}



{{{x = (4 +- sqrt( 76 ))/(2(3))}}} Add {{{16}}} to {{{60}}} to get {{{76}}}



{{{x = (4 +- sqrt( 76 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (4 +- 2*sqrt(19))/(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(19))/(6)}}} or {{{x = (4-2*sqrt(19))/(6)}}} Break up the expression.



{{{x = (2+sqrt(19))/(3)}}} or {{{x = (2-sqrt(19))/(3)}}} Reduce.  



So the solutions are {{{x = (2+sqrt(19))/(3)}}} or {{{x = (2-sqrt(19))/(3)}}}