4k^2+18k+18=0

Solution for 4k^2+18k+18=0 equation:

4k^2+18k+18=0

a = 4; b = 18; c = +18;
Δ = b2-4ac
Δ = 182-4·4·18
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*4}=\frac{-24}{8}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*4}=\frac{-12}{8}
=-1+1/2
$