Question 156773

{{{2x^2-5x=-9}}} Start with the given equation.



{{{2x^2-5x+9=0}}} Add 9 to both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=-5}}}, and {{{c=9}}}



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(2)(9) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-5}}}, and {{{c=9}}}



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



{{{x = (5 +- sqrt( 25-4(2)(9) ))/(2(2))}}} Square {{{-5}}} to get {{{25}}}. 



{{{x = (5 +- sqrt( 25-72 ))/(2(2))}}} Multiply {{{4(2)(9)}}} to get {{{72}}}



{{{x = (5 +- sqrt( -47 ))/(2(2))}}} Subtract {{{72}}} from {{{25}}} to get {{{-47}}}



{{{x = (5 +- sqrt( -47 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (5 +- i*sqrt(47))/(4)}}} Simplify the square root  



{{{x = (5+i*sqrt(47))/(4)}}} or {{{x = (5-i*sqrt(47))/(4)}}} Break up the expression.  



So the answers are {{{x = (5+i*sqrt(47))/(4)}}} or {{{x = (5-i*sqrt(47))/(4)}}} 



which approximate to {{{x=1.25+1.714*i}}} or {{{x=1.25+1.714*i}}}