-12k(k-11)+8(k+1)=1+9k-4

Solution for -12k(k-11)+8(k+1)=1+9k-4 equation:

-12k(k-11)+8(k+1)=1+9k-4

We move all terms to the left:

-12k(k-11)+8(k+1)-(1+9k-4)=0

We add all the numbers together, and all the variables

-12k(k-11)+8(k+1)-(9k-3)=0

We multiply parentheses

-12k^2+132k+8k-(9k-3)+8=0

We get rid of parentheses

-12k^2+132k+8k-9k+3+8=0

We add all the numbers together, and all the variables

-12k^2+131k+11=0

a = -12; b = 131; c = +11;
Δ = b2-4ac
Δ = 1312-4·(-12)·11
Δ = 17689
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{17689}=133$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(131)-133}{2*-12}=\frac{-264}{-24}
=+11
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(131)+133}{2*-12}=\frac{2}{-24}
=-1/12
$