Question 526361
To find the x-intercepts of {{{2x^2+5x-1}}}, we need to solve {{{2x^2+5x-1=0}}} for x.





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



Notice that the quadratic {{{2x^2+5x-1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=5}}}, 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 = (-(5) +- sqrt( (5)^2-4(2)(-1) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=5}}}, and {{{C=-1}}}



{{{x = (-5 +- sqrt( 25-4(2)(-1) ))/(2(2))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25--8 ))/(2(2))}}} Multiply {{{4(2)(-1)}}} to get {{{-8}}}



{{{x = (-5 +- sqrt( 25+8 ))/(2(2))}}} Rewrite {{{sqrt(25--8)}}} as {{{sqrt(25+8)}}}



{{{x = (-5 +- sqrt( 33 ))/(2(2))}}} Add {{{25}}} to {{{8}}} to get {{{33}}}



{{{x = (-5 +- sqrt( 33 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-5+sqrt(33))/(4)}}} or {{{x = (-5-sqrt(33))/(4)}}} Break up the expression.  



So the exact solutions are {{{x = (-5+sqrt(33))/(4)}}} or {{{x = (-5-sqrt(33))/(4)}}} 



which approximate to {{{x = 0.186}}} or {{{x = -2.686}}} (by using a calculator)



So the x-intercepts are (0.186, 0) and (-2.686, 0)