Question 453209


{{{8x^2+23x-3=0}}} Start with the given equation.



Notice that the quadratic {{{8x^2+23x-3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=8}}}, {{{B=23}}}, 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 = (-(23) +- sqrt( (23)^2-4(8)(-3) ))/(2(8))}}} Plug in  {{{A=8}}}, {{{B=23}}}, and {{{C=-3}}}



{{{x = (-23 +- sqrt( 529-4(8)(-3) ))/(2(8))}}} Square {{{23}}} to get {{{529}}}. 



{{{x = (-23 +- sqrt( 529--96 ))/(2(8))}}} Multiply {{{4(8)(-3)}}} to get {{{-96}}}



{{{x = (-23 +- sqrt( 529+96 ))/(2(8))}}} Rewrite {{{sqrt(529--96)}}} as {{{sqrt(529+96)}}}



{{{x = (-23 +- sqrt( 625 ))/(2(8))}}} Add {{{529}}} to {{{96}}} to get {{{625}}}



{{{x = (-23 +- sqrt( 625 ))/(16)}}} Multiply {{{2}}} and {{{8}}} to get {{{16}}}. 



{{{x = (-23 +- 25)/(16)}}} Take the square root of {{{625}}} to get {{{25}}}. 



{{{x = (-23 + 25)/(16)}}} or {{{x = (-23 - 25)/(16)}}} Break up the expression. 



{{{x = (2)/(16)}}} or {{{x =  (-48)/(16)}}} Combine like terms. 



{{{x = 1/8}}} or {{{x = -3}}} Simplify. 



So the solutions are {{{x = 1/8}}} or {{{x = -3}}}