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

Solution for 6(2k+3)=k(1+k) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

12k-(k(k+1))+18=0


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

k(k+1)

We multiply parentheses

k^2+k

Back to the equation:

-(k^2+k)


We get rid of parentheses

-k^2+12k-k+18=0

We add all the numbers together, and all the variables

-1k^2+11k+18=0

a = -1; b = 11; c = +18;
Δ = b2-4ac
Δ = 112-4·(-1)·18
Δ = 193
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{-(11)-\sqrt{193}}{2*-1}=\frac{-11-\sqrt{193}}{-2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{193}}{2*-1}=\frac{-11+\sqrt{193}}{-2}
$