Question 551828

{{{5x^2-4x-33=0}}} Start with the given equation.



Notice that the quadratic {{{5x^2-4x-33}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=5}}}, {{{B=-4}}}, and {{{C=-33}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-4) +- sqrt( (-4)^2-4(5)(-33) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=-4}}}, and {{{C=-33}}}



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



{{{x = (4 +- sqrt( 16-4(5)(-33) ))/(2(5))}}} Square {{{-4}}} to get {{{16}}}. 



{{{x = (4 +- sqrt( 16--660 ))/(2(5))}}} Multiply {{{4(5)(-33)}}} to get {{{-660}}}



{{{x = (4 +- sqrt( 16+660 ))/(2(5))}}} Rewrite {{{sqrt(16--660)}}} as {{{sqrt(16+660)}}}



{{{x = (4 +- sqrt( 676 ))/(2(5))}}} Add {{{16}}} to {{{660}}} to get {{{676}}}



{{{x = (4 +- sqrt( 676 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (4 +- 26)/(10)}}} Take the square root of {{{676}}} to get {{{26}}}. 



{{{x = (4 + 26)/(10)}}} or {{{x = (4 - 26)/(10)}}} Break up the expression. 



{{{x = (30)/(10)}}} or {{{x =  (-22)/(10)}}} Combine like terms. 



{{{x = 3}}} or {{{x = -11/5}}} Simplify. 



So the solutions are {{{x = 3}}} or {{{x = -11/5}}}