Question 591696


{{{4x^2-3x=-12}}} Start with the given equation.



{{{4x^2-3x+12=0}}} Get every term to the left side.



Notice that the quadratic {{{4x^2-3x+12}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=4}}}, {{{B=-3}}}, 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 = (-(-3) +- sqrt( (-3)^2-4(4)(12) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-3}}}, and {{{C=12}}}



{{{x = (3 +- sqrt( (-3)^2-4(4)(12) ))/(2(4))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(4)(12) ))/(2(4))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9-192 ))/(2(4))}}} Multiply {{{4(4)(12)}}} to get {{{192}}}



{{{x = (3 +- sqrt( -183 ))/(2(4))}}} Subtract {{{192}}} from {{{9}}} to get {{{-183}}}



{{{x = (3 +- sqrt( -183 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (3 +- i*sqrt(183))/(8)}}} 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(183))/(8)}}} or {{{x = (3-i*sqrt(183))/(8)}}} Break up the expression.  



So the solutions are {{{x = (3+i*sqrt(183))/(8)}}} or {{{x = (3-i*sqrt(183))/(8)}}}