Question 187590
I'm assuming that the equation is really {{{2x^2-7x-2=0}}}



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



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=-7}}}, and {{{c=-2}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-7) +- sqrt( (-7)^2-4(2)(-2) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-7}}}, and {{{c=-2}}}



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



{{{x = (7 +- sqrt( 49-4(2)(-2) ))/(2(2))}}} Square {{{-7}}} to get {{{49}}}. 



{{{x = (7 +- sqrt( 49--16 ))/(2(2))}}} Multiply {{{4(2)(-2)}}} to get {{{-16}}}



{{{x = (7 +- sqrt( 49+16 ))/(2(2))}}} Rewrite {{{sqrt(49--16)}}} as {{{sqrt(49+16)}}}



{{{x = (7 +- sqrt( 65 ))/(2(2))}}} Add {{{49}}} to {{{16}}} to get {{{65}}}



{{{x = (7 +- sqrt( 65 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (7+sqrt(65))/(4)}}} or {{{x = (7-sqrt(65))/(4)}}} Break up the expression.  



So the answers are {{{x = (7+sqrt(65))/(4)}}} or {{{x = (7-sqrt(65))/(4)}}}