Question

${x}^{2} + 4x + 4 = 0$

Answer 1

Δ= 4² - 4.1.4=
Δ= 16 - 16=
Δ=0

x= $\frac{-4+- \sqrt{0} }{2.1}$
x= $\frac{-4+-0}{2.1}$
x= $\frac{-4}{2}$
x= -2