K(x)=(x+4)(x+1)

Solution for K(x)=(x+4)(x+1) equation:

(K)=(K+4)(K+1)

We move all terms to the left:

(K)-((K+4)(K+1))=0

We multiply parentheses ..

-((+K^2+K+4K+4))+K=0


We calculate terms in parentheses: -((+K^2+K+4K+4)), so:

(+K^2+K+4K+4)

We get rid of parentheses

K^2+K+4K+4

We add all the numbers together, and all the variables

K^2+5K+4

Back to the equation:

-(K^2+5K+4)


We add all the numbers together, and all the variables

K-(K^2+5K+4)=0

We get rid of parentheses

-K^2+K-5K-4=0

We add all the numbers together, and all the variables

-1K^2-4K-4=0

a = -1; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·(-1)·(-4)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$K=\frac{-b}{2a}=\frac{4}{-2}=-2$