(k1+1)(2k+1)=15

Solution for (k1+1)(2k+1)=15 equation:

(k1+1)(2k+1)=15

We move all terms to the left:

(k1+1)(2k+1)-(15)=0

We add all the numbers together, and all the variables

(k+1)(2k+1)-15=0

We multiply parentheses ..

(+2k^2+k+2k+1)-15=0

We get rid of parentheses

2k^2+k+2k+1-15=0

We add all the numbers together, and all the variables

2k^2+3k-14=0

a = 2; b = 3; c = -14;
Δ = b2-4ac
Δ = 32-4·2·(-14)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*2}=\frac{-14}{4}
=-3+1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*2}=\frac{8}{4}
=2
$