Question 581832

{{{2x^2=5-3x}}} Start with the given equation.



{{{2x^2-5+3x=0}}} Get every term to the left side.



{{{2x^2+3x-5=0}}} Rearrange the terms.



Notice that the quadratic {{{2x^2+3x-5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=3}}}, and {{{C=-5}}}



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



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



{{{x = (-(3) +- sqrt( (3)^2-4(2)(-5) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=3}}}, and {{{C=-5}}}



{{{x = (-3 +- sqrt( 9-4(2)(-5) ))/(2(2))}}} Square {{{3}}} to get {{{9}}}. 



{{{x = (-3 +- sqrt( 9--40 ))/(2(2))}}} Multiply {{{4(2)(-5)}}} to get {{{-40}}}



{{{x = (-3 +- sqrt( 9+40 ))/(2(2))}}} Rewrite {{{sqrt(9--40)}}} as {{{sqrt(9+40)}}}



{{{x = (-3 +- sqrt( 49 ))/(2(2))}}} Add {{{9}}} to {{{40}}} to get {{{49}}}



{{{x = (-3 +- sqrt( 49 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-3 +- 7)/(4)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{x = (-3 + 7)/(4)}}} or {{{x = (-3 - 7)/(4)}}} Break up the expression. 



{{{x = (4)/(4)}}} or {{{x =  (-10)/(4)}}} Combine like terms. 



{{{x = 1}}} or {{{x = -5/2}}} Simplify. 



So the solutions are {{{x = 1}}} or {{{x = -5/2}}}