Question 334561
I'll do the first to get you started. Please only post one problem at a time.



{{{y^2+13y+22=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ay^2+by+c}}} where {{{a=1}}}, {{{b=13}}}, and {{{c=22}}}



Let's use the quadratic formula to solve for y



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



{{{y = (-(13) +- sqrt( (13)^2-4(1)(22) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=13}}}, and {{{c=22}}}



{{{y = (-13 +- sqrt( 169-4(1)(22) ))/(2(1))}}} Square {{{13}}} to get {{{169}}}. 



{{{y = (-13 +- sqrt( 169-88 ))/(2(1))}}} Multiply {{{4(1)(22)}}} to get {{{88}}}



{{{y = (-13 +- sqrt( 81 ))/(2(1))}}} Subtract {{{88}}} from {{{169}}} to get {{{81}}}



{{{y = (-13 +- sqrt( 81 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{y = (-13 +- 9)/(2)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{y = (-13 + 9)/(2)}}} or {{{y = (-13 - 9)/(2)}}} Break up the expression. 



{{{y = (-4)/(2)}}} or {{{y =  (-22)/(2)}}} Combine like terms. 



{{{y = -2}}} or {{{y = -11}}} Simplify. 



So the answers are {{{y = -2}}} or {{{y = -11}}}