Question 236208


{{{4y^2-7y-15=0}}} Start with the given equation.



Notice that the quadratic {{{4y^2-7y-15}}} is in the form of {{{Ay^2+By+C}}} where {{{A=4}}}, {{{B=-7}}}, and {{{C=-15}}}



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



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



{{{y = (-(-7) +- sqrt( (-7)^2-4(4)(-15) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-7}}}, and {{{C=-15}}}



{{{y = (7 +- sqrt( (-7)^2-4(4)(-15) ))/(2(4))}}} Negate {{{-7}}} to get {{{7}}}. 



{{{y = (7 +- sqrt( 49-4(4)(-15) ))/(2(4))}}} Square {{{-7}}} to get {{{49}}}. 



{{{y = (7 +- sqrt( 49--240 ))/(2(4))}}} Multiply {{{4(4)(-15)}}} to get {{{-240}}}



{{{y = (7 +- sqrt( 49+240 ))/(2(4))}}} Rewrite {{{sqrt(49--240)}}} as {{{sqrt(49+240)}}}



{{{y = (7 +- sqrt( 289 ))/(2(4))}}} Add {{{49}}} to {{{240}}} to get {{{289}}}



{{{y = (7 +- sqrt( 289 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{y = (7 +- 17)/(8)}}} Take the square root of {{{289}}} to get {{{17}}}. 



{{{y = (7 + 17)/(8)}}} or {{{y = (7 - 17)/(8)}}} Break up the expression. 



{{{y = (24)/(8)}}} or {{{y =  (-10)/(8)}}} Combine like terms. 



{{{y = 3}}} or {{{y = -5/4}}} Simplify. 



So the solutions are {{{y = 3}}} or {{{y = -5/4}}}