Question 407319
{{{2x^2+6x=3}}} Start with the given equation.



{{{2x^2+6x-3=0}}} Subtract 3 from both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=6}}}, 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 = (-(6) +- sqrt( (6)^2-4(2)(-3) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=6}}}, and {{{c=-3}}}



{{{x = (-6 +- sqrt( 36-4(2)(-3) ))/(2(2))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36--24 ))/(2(2))}}} Multiply {{{4(2)(-3)}}} to get {{{-24}}}



{{{x = (-6 +- sqrt( 36+24 ))/(2(2))}}} Rewrite {{{sqrt(36--24)}}} as {{{sqrt(36+24)}}}



{{{x = (-6 +- sqrt( 60 ))/(2(2))}}} Add {{{36}}} to {{{24}}} to get {{{60}}}



{{{x = (-6 +- sqrt( 60 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-6 +- 2*sqrt(15))/(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 = (-6+2*sqrt(15))/(4)}}} or {{{x = (-6-2*sqrt(15))/(4)}}} Break up the expression.  



{{{x = (-3+sqrt(15))/(2)}}} or {{{x = (-3-sqrt(15))/(2)}}} Reduce



So the answers are {{{x = (-3+sqrt(15))/(2)}}} or {{{x = (-3-sqrt(15))/(2)}}} 



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