5-6K-1=(5k-3)-(k-2)(5k-3)-(k-2)

Solution for 5-6K-1=(5k-3)-(k-2)(5k-3)-(k-2) equation:

5-6-1=(5K-3)-(K-2)(5K-3)-(K-2)

We move all terms to the left:

5-6-1-((5K-3)-(K-2)(5K-3)-(K-2))=0

We add all the numbers together, and all the variables

-((5K-3)-(K-2)(5K-3)-(K-2))-2=0

We multiply parentheses ..

-((5K-3)-(+5K^2-3K-10K+6)-(K-2))-2=0


We calculate terms in parentheses: -((5K-3)-(+5K^2-3K-10K+6)-(K-2)), so:

(5K-3)-(+5K^2-3K-10K+6)-(K-2)

determiningTheFunctionDomain
-(+5K^2-3K-10K+6)+(5K-3)-(K-2)

We get rid of parentheses

-5K^2+3K+10K+5K-K-6-3+2

We add all the numbers together, and all the variables

-5K^2+17K-7

Back to the equation:

-(-5K^2+17K-7)


We get rid of parentheses

5K^2-17K+7-2=0

We add all the numbers together, and all the variables

5K^2-17K+5=0

a = 5; b = -17; c = +5;
Δ = b2-4ac
Δ = -172-4·5·5
Δ = 189
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-3\sqrt{21}}{2*5}=\frac{17-3\sqrt{21}}{10}
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+3\sqrt{21}}{2*5}=\frac{17+3\sqrt{21}}{10}
$