(2k-1)(2k+1)=k^2+(2k+1)

Solution for (2k-1)(2k+1)=k^2+(2k+1) equation:

(2k-1)(2k+1)=k^2+(2k+1)

We move all terms to the left:

(2k-1)(2k+1)-(k^2+(2k+1))=0

We use the square of the difference formula

4k^2-(k^2+(2k+1))-1=0


We calculate terms in parentheses: -(k^2+(2k+1)), so:

k^2+(2k+1)

We get rid of parentheses

k^2+2k+1

Back to the equation:

-(k^2+2k+1)


We get rid of parentheses

4k^2-k^2-2k-1-1=0

We add all the numbers together, and all the variables

3k^2-2k-2=0

a = 3; b = -2; c = -2;
Δ = b2-4ac
Δ = -22-4·3·(-2)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*3}=\frac{2-2\sqrt{7}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*3}=\frac{2+2\sqrt{7}}{6}
$