Question 189892


{{{-2h^2-28h-98=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ah^2+bh+c}}} where {{{a=-2}}}, {{{b=-28}}}, and {{{c=-98}}}



Let's use the quadratic formula to solve for h



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



{{{h = (-(-28) +- sqrt( (-28)^2-4(-2)(-98) ))/(2(-2))}}} Plug in  {{{a=-2}}}, {{{b=-28}}}, and {{{c=-98}}}



{{{h = (28 +- sqrt( (-28)^2-4(-2)(-98) ))/(2(-2))}}} Negate {{{-28}}} to get {{{28}}}. 



{{{h = (28 +- sqrt( 784-4(-2)(-98) ))/(2(-2))}}} Square {{{-28}}} to get {{{784}}}. 



{{{h = (28 +- sqrt( 784-784 ))/(2(-2))}}} Multiply {{{4(-2)(-98)}}} to get {{{784}}}



{{{h = (28 +- sqrt( 0 ))/(2(-2))}}} Subtract {{{784}}} from {{{784}}} to get {{{0}}}



{{{h = (28 +- sqrt( 0 ))/(-4)}}} Multiply {{{2}}} and {{{-2}}} to get {{{-4}}}. 



{{{h = (28 +- 0)/(-4)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{h = (28 + 0)/(-4)}}} or {{{h = (28 - 0)/(-4)}}} Break up the expression. 



{{{h = (28)/(-4)}}} or {{{h =  (28)/(-4)}}} Combine like terms. 



{{{h = -7}}} or {{{h = -7}}} Simplify. 



So the answer is {{{h = -7}}}