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

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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} $

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