What are the solutions to this equation: x^2+4x+4=-5?

x²+4x+4= -5
Since the LHS is already a perfect square, your work is very easy
x²+4x+4= -5
(x + 2)² = -5.................rewrite the LHS as a square
(x + 2) = ± √-5.............extract squareroots
(x + 2) = ± i√5.............I hope you understand this
x = - 2 ± i√5................Transpose the 2 and you have the final answer.

x² + 4x + 4 = - 5
x² + 2x = - 9 + 2²
x² + 2x = - 9 + 4
(x + 2)² = - 5
x + 2 = 5i

x = 5i - 2
x = - 5i - 2

Answer: x = (5i - 2), - (5i + 2)

x^2 + 4x + 4 + 5 = 0
x^2 + 4x + 9 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = 4
c = 9

x = [-4 ±√(16 - 36)]/2
x = [-4 ±√(-20)]/2
x = [-4 ±i√(2^2 * 5)]/2
x = [-4 ±2i√5]/2
x = -2 ±i√5

∴ x = -2 ±i√5

x² + 4x + 9 = 0
x = [ - b ± √ (b ² - 4 a c ) ] / 2 a
x = [ - 4 ± √ (16 - 36 ) ] / 2
x = [ - 4 ± √ (- 20) ] / 2
x = [ - 4 ± √ ( 20 i ² ) ] / 2
x = [ - 4 ± i 2 √5 ] / 2
x = - 2 ± i √5

You were close. Edit: nvm you got it.

x^2+4x+4=-5
(x+2)^2=-5
x+2=±√(-5)
x=-2±√[(5)(-1)]
x=-2±i√(5)

Note: I used Alt241 for ± and Alt 251 for √.

You could use the Quadratic Formula, but I just noticed the left side is factorable. Use that.

(x + 2)^2 = -5

Because the left side is a perfect square.

Take square root of both sides.

x + 2 = sqrt( -5)

Of course, taking the square root of a negative number gets you into the realm of imaginary numbers, right? Remember that i = sqrt(-1).

Thus, sqrt(-5) = i*sqrt(5).

Now, since a*a = (-a)(-a) = a^2, we say that the square root of a^2 is +/-a.

Now just assemble the parts and move the 2 over to the right side.

BTW, it should be -2 +/- i[rad(5)].

I'm not doing your homework.