(4k+4)(2k+4)=0

Solution for (4k+4)(2k+4)=0 equation:

(4k+4)(2k+4)=0

We multiply parentheses ..

(+8k^2+16k+8k+16)=0

We get rid of parentheses

8k^2+16k+8k+16=0

We add all the numbers together, and all the variables

8k^2+24k+16=0

a = 8; b = 24; c = +16;
Δ = b2-4ac
Δ = 242-4·8·16
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8}{2*8}=\frac{-32}{16}
=-2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8}{2*8}=\frac{-16}{16}
=-1
$