Question 592834
{{{ -5x^2= 12-3x }}} Start with the given equation.



{{{ 0= 12-3x+5x^2 }}} Add {{{5x^2}}} to both sides



{{{0 = 5x^2-3x+12}}} Rearrange the terms.



{{{5x^2-3x+12 = 0}}} Flip the equation.



Notice that the quadratic {{{5x^2-3x+12}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=5}}}, {{{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(5)(12) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=-3}}}, and {{{C=12}}}



{{{x = (3 +- sqrt( (-3)^2-4(5)(12) ))/(2(5))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(5)(12) ))/(2(5))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9-240 ))/(2(5))}}} Multiply {{{4(5)(12)}}} to get {{{240}}}



{{{x = (3 +- sqrt( -231 ))/(2(5))}}} Subtract {{{240}}} from {{{9}}} to get {{{-231}}}



{{{x = (3 +- sqrt( -231 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (3 +- i*sqrt(231))/(10)}}} 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(231))/(10)}}} or {{{x = (3-i*sqrt(231))/(10)}}} Break up the expression.  



So the solutions are {{{x = (3+i*sqrt(231))/(10)}}} or {{{x = (3-i*sqrt(231))/(10)}}}