```Question 266445

{{{y^2+4y+4-7=0}}} Subtract 7 from both sides.

{{{y^2+4y-3=0}}} Combine like terms.

Notice that the quadratic {{{y^2+4y-3}}} is in the form of {{{Ay^2+By+C}}} where {{{A=1}}}, {{{B=4}}}, and {{{C=-3}}}

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

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

{{{y = (-4 +- sqrt( 16-4(1)(-3) ))/(2(1))}}} Square {{{4}}} to get {{{16}}}.

{{{y = (-4 +- sqrt( 16--12 ))/(2(1))}}} Multiply {{{4(1)(-3)}}} to get {{{-12}}}

{{{y = (-4 +- sqrt( 16+12 ))/(2(1))}}} Rewrite {{{sqrt(16--12)}}} as {{{sqrt(16+12)}}}

{{{y = (-4 +- sqrt( 28 ))/(2(1))}}} Add {{{16}}} to {{{12}}} to get {{{28}}}

{{{y = (-4 +- sqrt( 28 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}.

{{{y = (-4 +- 2*sqrt(7))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)

{{{y = (-4)/(2) +- (2*sqrt(7))/(2)}}} Break up the fraction.

{{{y = -2 +- sqrt(7)}}} Reduce.

{{{y = -2+sqrt(7)}}} or {{{y = -2-sqrt(7)}}} Break up the expression.

So the solutions are {{{y = -2+sqrt(7)}}} or {{{y = -2-sqrt(7)}}} ```