Question 147559


{{{2x^2-x=-10}}} Start with the given equation.



{{{2x^2-x+10=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=-1}}}, and {{{c=10}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-1) +- sqrt( (-1)^2-4(2)(10) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-1}}}, and {{{c=10}}}



{{{x = (1 +- sqrt( (-1)^2-4(2)(10) ))/(2(2))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(2)(10) ))/(2(2))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-80 ))/(2(2))}}} Multiply {{{4(2)(10)}}} to get {{{80}}}



{{{x = (1 +- sqrt( -79 ))/(2(2))}}} Subtract {{{80}}} from {{{1}}} to get {{{-79}}}



{{{x = (1 +- sqrt( -79 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (1 +- i*sqrt(79))/(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 = (1+i*sqrt(79))/(4)}}} or {{{x = (1-i*sqrt(79))/(4)}}} Break up the expression.  



So our answers are {{{x = (1+i*sqrt(79))/(4)}}} or {{{x = (1-i*sqrt(79))/(4)}}} 



which approximate to {{{x=2.472*i}}} or {{{x=-1.972*i}}}