Question 553059


{{{2x^2-x=12-x}}} Start with the given equation.



{{{2x^2-x-12+1x=0}}} Get every term to the left side.



{{{2x^2+0x-12=0}}} Combine like terms.



Notice that the quadratic {{{2x^2+0x-12}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=0}}}, and {{{C=-12}}}



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



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



{{{x = (-(0) +- sqrt( (0)^2-4(2)(-12) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=0}}}, and {{{C=-12}}}



{{{x = (-0 +- sqrt( 0-4(2)(-12) ))/(2(2))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (-0 +- sqrt( 0--96 ))/(2(2))}}} Multiply {{{4(2)(-12)}}} to get {{{-96}}}



{{{x = (-0 +- sqrt( 0+96 ))/(2(2))}}} Rewrite {{{sqrt(0--96)}}} as {{{sqrt(0+96)}}}



{{{x = (-0 +- sqrt( 96 ))/(2(2))}}} Add {{{0}}} to {{{96}}} to get {{{96}}}



{{{x = (-0 +- sqrt( 96 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-0 +- 4*sqrt(6))/(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 = (-0)/(4) +- (4*sqrt(6))/(4)}}} Break up the fraction.  



{{{x = 0 +- sqrt(6)}}} Reduce.  



{{{x = sqrt(6)}}} or {{{x = -sqrt(6)}}} Break up the expression.  



So the solutions are {{{x = sqrt(6)}}} or {{{x = -sqrt(6)}}}