Question 275262
The zeros are the x values that make {{{f(x)=0}}}. In other words, they are the x values that satisfy the equation {{{5x^2 - 2x - 1=0}}}





{{{5x^2-2x-1=0}}} Start with the given equation.



Notice that the quadratic {{{5x^2-2x-1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=5}}}, {{{B=-2}}}, and {{{C=-1}}}



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(5)(-1) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=-2}}}, and {{{C=-1}}}



{{{x = (2 +- sqrt( (-2)^2-4(5)(-1) ))/(2(5))}}} Negate {{{-2}}} to get {{{2}}}. 



{{{x = (2 +- sqrt( 4-4(5)(-1) ))/(2(5))}}} Square {{{-2}}} to get {{{4}}}. 



{{{x = (2 +- sqrt( 4--20 ))/(2(5))}}} Multiply {{{4(5)(-1)}}} to get {{{-20}}}



{{{x = (2 +- sqrt( 4+20 ))/(2(5))}}} Rewrite {{{sqrt(4--20)}}} as {{{sqrt(4+20)}}}



{{{x = (2 +- sqrt( 24 ))/(2(5))}}} Add {{{4}}} to {{{20}}} to get {{{24}}}



{{{x = (2 +- sqrt( 24 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (2 +- 2*sqrt(6))/(10)}}} 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(6))/(10)}}} or {{{x = (2-2*sqrt(6))/(10)}}} Break up the expression.  



{{{x = (1+sqrt(6))/(5)}}} or {{{x = (1-sqrt(6))/(5)}}} Reduce.



So the solutions are {{{x = (1+sqrt(6))/(5)}}} or {{{x = (1-sqrt(6))/(5)}}}