K2+5k+6=(k+2)(k+)

Solution for K2+5k+6=(k+2)(k+) equation:

2+5K+6=(K+2)(K+)

We move all terms to the left:

2+5K+6-((K+2)(K+))=0

We add all the numbers together, and all the variables

5K-((K+2)(+K))+2+6=0

We add all the numbers together, and all the variables

5K-((K+2)(+K))+8=0

We multiply parentheses ..

-((+K^2+2K))+5K+8=0


We calculate terms in parentheses: -((+K^2+2K)), so:

(+K^2+2K)

We get rid of parentheses

K^2+2K

Back to the equation:

-(K^2+2K)


We add all the numbers together, and all the variables

5K-(K^2+2K)+8=0

We get rid of parentheses

-K^2+5K-2K+8=0

We add all the numbers together, and all the variables

-1K^2+3K+8=0

a = -1; b = 3; c = +8;
Δ = b2-4ac
Δ = 32-4·(-1)·8
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{41}}{2*-1}=\frac{-3-\sqrt{41}}{-2}
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*-1}=\frac{-3+\sqrt{41}}{-2}
$