Question 405271
You're doing great until you come onto the step where you factor. So I'm going to start with the step before that one.




{{{y^2-14=5y }}} Start with the given equation.



{{{y^2-5y-14=0 }}} Subtract 5y from both sides.



Notice we have a quadratic equation in the form of {{{ay^2+by+c}}} where {{{a=1}}}, {{{b=-5}}}, and {{{c=-14}}}



Let's use the quadratic formula to solve for y



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



{{{y = (-(-5) +- sqrt( (-5)^2-4(1)(-14) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-5}}}, and {{{c=-14}}}



{{{y = (5 +- sqrt( (-5)^2-4(1)(-14) ))/(2(1))}}} Negate {{{-5}}} to get {{{5}}}. 



{{{y = (5 +- sqrt( 25-4(1)(-14) ))/(2(1))}}} Square {{{-5}}} to get {{{25}}}. 



{{{y = (5 +- sqrt( 25--56 ))/(2(1))}}} Multiply {{{4(1)(-14)}}} to get {{{-56}}}



{{{y = (5 +- sqrt( 25+56 ))/(2(1))}}} Rewrite {{{sqrt(25--56)}}} as {{{sqrt(25+56)}}}



{{{y = (5 +- sqrt( 81 ))/(2(1))}}} Add {{{25}}} to {{{56}}} to get {{{81}}}



{{{y = (5 +- sqrt( 81 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{y = (5 +- 9)/(2)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{y = (5 + 9)/(2)}}} or {{{y = (5 - 9)/(2)}}} Break up the expression. 



{{{y = (14)/(2)}}} or {{{y =  (-4)/(2)}}} Combine like terms. 



{{{y = 7}}} or {{{y = -2}}} Simplify. 



So the answers are {{{y = 7}}} or {{{y = -2}}} 



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