Question 201194


{{{3m^2+12m+5=0}}} Start with the given equation.



Notice that the quadratic {{{3m^2+12m+5}}} is in the form of {{{Am^2+Bm+C}}} where {{{A=3}}}, {{{B=12}}}, and {{{C=5}}}



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



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



{{{m = (-(12) +- sqrt( (12)^2-4(3)(5) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=12}}}, and {{{C=5}}}



{{{m = (-12 +- sqrt( 144-4(3)(5) ))/(2(3))}}} Square {{{12}}} to get {{{144}}}. 



{{{m = (-12 +- sqrt( 144-60 ))/(2(3))}}} Multiply {{{4(3)(5)}}} to get {{{60}}}



{{{m = (-12 +- sqrt( 84 ))/(2(3))}}} Subtract {{{60}}} from {{{144}}} to get {{{84}}}



{{{m = (-12 +- sqrt( 84 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{m = (-12 +- 2*sqrt(21))/(6)}}} 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>)  



{{{m = (-12+2*sqrt(21))/(6)}}} or {{{m = (-12-2*sqrt(21))/(6)}}} Break up the expression.  



{{{m = (-6+sqrt(21))/(3)}}} or {{{m = (-6-sqrt(21))/(3)}}} Reduce.



So the solutions are {{{m = (-6+sqrt(21))/(3)}}} or {{{m = (-6-sqrt(21))/(3)}}} 



which approximate to {{{m=-0.472}}} or {{{m=-3.528}}}