Question 207139

{{{7x^2+5x-2=0}}} Start with the given equation.



Notice that the quadratic {{{7x^2+5x-2}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=7}}}, {{{B=5}}}, 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 = (-(5) +- sqrt( (5)^2-4(7)(-2) ))/(2(7))}}} Plug in  {{{A=7}}}, {{{B=5}}}, and {{{C=-2}}}



{{{x = (-5 +- sqrt( 25-4(7)(-2) ))/(2(7))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25--56 ))/(2(7))}}} Multiply {{{4(7)(-2)}}} to get {{{-56}}}



{{{x = (-5 +- sqrt( 25+56 ))/(2(7))}}} Rewrite {{{sqrt(25--56)}}} as {{{sqrt(25+56)}}}



{{{x = (-5 +- sqrt( 81 ))/(2(7))}}} Add {{{25}}} to {{{56}}} to get {{{81}}}



{{{x = (-5 +- sqrt( 81 ))/(14)}}} Multiply {{{2}}} and {{{7}}} to get {{{14}}}. 



{{{x = (-5 +- 9)/(14)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{x = (-5 + 9)/(14)}}} or {{{x = (-5 - 9)/(14)}}} Break up the expression. 



{{{x = (4)/(14)}}} or {{{x =  (-14)/(14)}}} Combine like terms. 



{{{x = 2/7}}} or {{{x = -1}}} Simplify. 



So the solutions are {{{x = 2/7}}} or {{{x = -1}}}