Question 169982
{{{1 = 3q^2 + 4q}}} Start with the given equation



{{{0 = 3q^2 + 4q-1}}} Subtract 1 from both sides.



Notice we have a quadratic equation in the form of {{{aq^2+bq+c}}} where {{{a=3}}}, {{{b=4}}}, and {{{c=-1}}}



Let's use the quadratic formula to solve for q



{{{q = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{q = (-(4) +- sqrt( (4)^2-4(3)(-1) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=4}}}, and {{{c=-1}}}



{{{q = (-4 +- sqrt( 16-4(3)(-1) ))/(2(3))}}} Square {{{4}}} to get {{{16}}}. 



{{{q = (-4 +- sqrt( 16--12 ))/(2(3))}}} Multiply {{{4(3)(-1)}}} to get {{{-12}}}



{{{q = (-4 +- sqrt( 16+12 ))/(2(3))}}} Rewrite {{{sqrt(16--12)}}} as {{{sqrt(16+12)}}}



{{{q = (-4 +- sqrt( 28 ))/(2(3))}}} Add {{{16}}} to {{{12}}} to get {{{28}}}



{{{q = (-4 +- sqrt( 28 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{q = (-4 +- 2*sqrt(7))/(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>)  



{{{q = (-4+2*sqrt(7))/(6)}}} or {{{q = (-4-2*sqrt(7))/(6)}}} Break up the expression.  



So the answers are {{{q = (-4+2*sqrt(7))/(6)}}} or {{{q = (-4-2*sqrt(7))/(6)}}} 



which approximate to {{{q=0.215}}} or {{{q=-1.549}}}