Question 315573


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



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



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



{{{x = (3 +- sqrt( 9-4(7)(-3) ))/(2(7))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9--84 ))/(2(7))}}} Multiply {{{4(7)(-3)}}} to get {{{-84}}}



{{{x = (3 +- sqrt( 9+84 ))/(2(7))}}} Rewrite {{{sqrt(9--84)}}} as {{{sqrt(9+84)}}}



{{{x = (3 +- sqrt( 93 ))/(2(7))}}} Add {{{9}}} to {{{84}}} to get {{{93}}}



{{{x = (3 +- sqrt( 93 ))/(14)}}} Multiply {{{2}}} and {{{7}}} to get {{{14}}}. 



{{{x = (3+sqrt(93))/(14)}}} or {{{x = (3-sqrt(93))/(14)}}} Break up the expression.  



So the solutions are {{{x = (3+sqrt(93))/(14)}}} or {{{x = (3-sqrt(93))/(14)}}}