Question 147088


{{{2x^2-3x=-6}}} Start with the given equation.



{{{2x^2-3x+6=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=6}}}



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)(6) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-3}}}, and {{{c=6}}}



{{{x = (3 +- sqrt( (-3)^2-4(2)(6) ))/(2(2))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(2)(6) ))/(2(2))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9-48 ))/(2(2))}}} Multiply {{{4(2)(6)}}} to get {{{48}}}



{{{x = (3 +- sqrt( -39 ))/(2(2))}}} Subtract {{{48}}} from {{{9}}} to get {{{-39}}}



{{{x = (3 +- sqrt( -39 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (3 +- i*sqrt(39))/(4)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (3+i*sqrt(39))/(4)}}} or {{{x = (3-i*sqrt(39))/(4)}}} Break up the expression.  



So our answers are {{{x = (3+i*sqrt(39))/(4)}}} or {{{x = (3-i*sqrt(39))/(4)}}} 



which approximate to {{{x=2.311*i}}} or {{{x=-0.811*i}}}