(k+1)(2k+1)=5

Solution for (k+1)(2k+1)=5 equation:

(k+1)(2k+1)=5

We move all terms to the left:

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

We multiply parentheses ..

(+2k^2+k+2k+1)-5=0

We get rid of parentheses

2k^2+k+2k+1-5=0

We add all the numbers together, and all the variables

2k^2+3k-4=0

a = 2; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·2·(-4)
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*2}=\frac{-3-\sqrt{41}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*2}=\frac{-3+\sqrt{41}}{4}
$