K1^3+6k^2+9k+4=

Solution for K1^3+6k^2+9k+4= equation:

1^3+6K^2+9K+4=

We move all terms to the left:

1^3+6K^2+9K+4-()=0

We add all the numbers together, and all the variables

6K^2+9K=0

a = 6; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·6·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*6}=\frac{-18}{12}
=-1+1/2
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*6}=\frac{0}{12}
=0
$