Question 592280


{{{3h^2+2h-16=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ah^2+bh+c}}} where {{{a=3}}}, {{{b=2}}}, and {{{c=-16}}}



Let's use the quadratic formula to solve for h



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



{{{h = (-(2) +- sqrt( (2)^2-4(3)(-16) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=2}}}, and {{{c=-16}}}



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



{{{h = (-2 +- sqrt( 4--192 ))/(2(3))}}} Multiply {{{4(3)(-16)}}} to get {{{-192}}}



{{{h = (-2 +- sqrt( 4+192 ))/(2(3))}}} Rewrite {{{sqrt(4--192)}}} as {{{sqrt(4+192)}}}



{{{h = (-2 +- sqrt( 196 ))/(2(3))}}} Add {{{4}}} to {{{192}}} to get {{{196}}}



{{{h = (-2 +- sqrt( 196 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{h = (-2 +- 14)/(6)}}} Take the square root of {{{196}}} to get {{{14}}}. 



{{{h = (-2 + 14)/(6)}}} or {{{h = (-2 - 14)/(6)}}} Break up the expression. 



{{{h = (12)/(6)}}} or {{{h =  (-16)/(6)}}} Combine like terms. 



{{{h = 2}}} or {{{h = -8/3}}} Simplify. 



So the answers are {{{h = 2}}} or {{{h = -8/3}}}