Question 322828


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



{{{2x^2-4x-3=0}}} Subtract 3 from both sides.



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



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



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



{{{x = (-(-4) +- sqrt( (-4)^2-4(2)(-3) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=-4}}}, and {{{C=-3}}}



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



{{{x = (4 +- sqrt( 16-4(2)(-3) ))/(2(2))}}} Square {{{-4}}} to get {{{16}}}. 



{{{x = (4 +- sqrt( 16--24 ))/(2(2))}}} Multiply {{{4(2)(-3)}}} to get {{{-24}}}



{{{x = (4 +- sqrt( 16+24 ))/(2(2))}}} Rewrite {{{sqrt(16--24)}}} as {{{sqrt(16+24)}}}



{{{x = (4 +- sqrt( 40 ))/(2(2))}}} Add {{{16}}} to {{{24}}} to get {{{40}}}



{{{x = (4 +- sqrt( 40 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (4 +- 2*sqrt(10))/(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 = (4+2*sqrt(10))/(4)}}} or {{{x = (4-2*sqrt(10))/(4)}}} Break up the expression.  



{{{x = (2+sqrt(10))/(2)}}} or {{{x = (2-sqrt(10))/(2)}}} Reduce.



So the solutions are {{{x = (2+sqrt(10))/(2)}}} or {{{x = (2-sqrt(10))/(2)}}}