Question 188693

{{{3x^2+6=18}}} Start with the given equation.



{{{3x^2+6-18=0}}} Subtract 18 from both sides.



{{{3x^2-12=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=3}}}, {{{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(3)(-12) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=0}}}, and {{{c=-12}}}



{{{x = (0 +- sqrt( 0-4(3)(-12) ))/(2(3))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (0 +- sqrt( 0--144 ))/(2(3))}}} Multiply {{{4(3)(-12)}}} to get {{{-144}}}



{{{x = (0 +- sqrt( 0+144 ))/(2(3))}}} Rewrite {{{sqrt(0--144)}}} as {{{sqrt(0+144)}}}



{{{x = (0 +- sqrt( 144 ))/(2(3))}}} Add {{{0}}} to {{{144}}} to get {{{144}}}



{{{x = (0 +- sqrt( 144 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (0 +- 12)/(6)}}} Take the square root of {{{144}}} to get {{{12}}}. 



{{{x = (0 + 12)/(6)}}} or {{{x = (-0 - 12)/(6)}}} Break up the expression. 



{{{x = (12)/(6)}}} or {{{x =  (-12)/(6)}}} Combine like terms. 



{{{x = 2}}} or {{{x = -2}}} Simplify. 



So the answers are {{{x = 2}}} or {{{x = -2}}}




However, since {{{x<0}}}, this means that "x" is negative. So this means that {{{x=-2}}} is the only solution (if the inequality restriction is applied).