Question 467352


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



Notice that the quadratic {{{5x^2-3x-3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=5}}}, {{{B=-3}}}, 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 = (-(-3) +- sqrt( (-3)^2-4(5)(-3) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=-3}}}, and {{{C=-3}}}



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



{{{x = (3 +- sqrt( 9-4(5)(-3) ))/(2(5))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9--60 ))/(2(5))}}} Multiply {{{4(5)(-3)}}} to get {{{-60}}}



{{{x = (3 +- sqrt( 9+60 ))/(2(5))}}} Rewrite {{{sqrt(9--60)}}} as {{{sqrt(9+60)}}}



{{{x = (3 +- sqrt( 69 ))/(2(5))}}} Add {{{9}}} to {{{60}}} to get {{{69}}}



{{{x = (3 +- sqrt( 69 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (3+sqrt(69))/(10)}}} or {{{x = (3-sqrt(69))/(10)}}} Break up the expression.  



So the solutions are {{{x = (3+sqrt(69))/(10)}}} or {{{x = (3-sqrt(69))/(10)}}}