Question 147128


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



{{{2x^2-3x-3=0}}} Get all terms to the left side.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=-3}}}, 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 = (-(-3) +- sqrt( (-3)^2-4(2)(-3) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-3}}}, and {{{c=-3}}}



{{{x = (3 +- sqrt( (-3)^2-4(2)(-3) ))/(2(2))}}} Negate {{{-3}}} to get {{{3}}}. 



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



{{{x = (3 +- sqrt( 9--24 ))/(2(2))}}} Multiply {{{4(2)(-3)}}} to get {{{-24}}}



{{{x = (3 +- sqrt( 9+24 ))/(2(2))}}} Rewrite {{{sqrt(9--24)}}} as {{{sqrt(9+24)}}}



{{{x = (3 +- sqrt( 33 ))/(2(2))}}} Add {{{9}}} to {{{24}}} to get {{{33}}}



{{{x = (3 +- sqrt( 33 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (3+sqrt(33))/(4)}}} or {{{x = (3-sqrt(33))/(4)}}} Break up the expression.  



So our answers are {{{x = (3+sqrt(33))/(4)}}} or {{{x = (3-sqrt(33))/(4)}}} 



which approximate to {{{x=2.186}}} or {{{x=-0.686}}}