Question 200756

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



Notice that the quadratic {{{2h^2-h-3}}} is in the form of {{{Ah^2+Bh+C}}} where {{{A=2}}}, {{{B=-1}}}, and {{{C=-3}}}



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



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



{{{h = (-(-1) +- sqrt( (-1)^2-4(2)(-3) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=-1}}}, and {{{C=-3}}}



{{{h = (1 +- sqrt( (-1)^2-4(2)(-3) ))/(2(2))}}} Negate {{{-1}}} to get {{{1}}}. 



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



{{{h = (1 +- sqrt( 1--24 ))/(2(2))}}} Multiply {{{4(2)(-3)}}} to get {{{-24}}}



{{{h = (1 +- sqrt( 1+24 ))/(2(2))}}} Rewrite {{{sqrt(1--24)}}} as {{{sqrt(1+24)}}}



{{{h = (1 +- sqrt( 25 ))/(2(2))}}} Add {{{1}}} to {{{24}}} to get {{{25}}}



{{{h = (1 +- sqrt( 25 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{h = (1 +- 5)/(4)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{h = (1 + 5)/(4)}}} or {{{h = (1 - 5)/(4)}}} Break up the expression. 



{{{h = (6)/(4)}}} or {{{h =  (-4)/(4)}}} Combine like terms. 



{{{h = 3/2}}} or {{{h = -1}}} Simplify. 



So the solutions are {{{h = 3/2}}} or {{{h = -1}}}