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

We move all terms to the left:

5-(6k+1)-((5k-3)-(k-2)(5k-3)-(k-2))=0

We get rid of parentheses

-6k-((5k-3)-(k-2)(5k-3)-(k-2))-1+5=0

We multiply parentheses ..

-((5k-3)-(+5k^2-3k-10k+6)-(k-2))-6k-1+5=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 add all the numbers together, and all the variables

-(-5k^2+17k-7)-6k+4=0

We get rid of parentheses

5k^2-17k-6k+7+4=0

We add all the numbers together, and all the variables

5k^2-23k+11=0

a = 5; b = -23; c = +11;
Δ = b2-4ac
Δ = -232-4·5·11
Δ = 309
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{-(-23)-\sqrt{309}}{2*5}=\frac{23-\sqrt{309}}{10}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{309}}{2*5}=\frac{23+\sqrt{309}}{10}
$