3k(k-4)+k+4(k+1)=0

Solution for 3k(k-4)+k+4(k+1)=0 equation:

3k(k-4)+k+4(k+1)=0

We add all the numbers together, and all the variables

k+3k(k-4)+4(k+1)=0

We multiply parentheses

3k^2+k-12k+4k+4=0

We add all the numbers together, and all the variables

3k^2-7k+4=0

a = 3; b = -7; c = +4;
Δ = b2-4ac
Δ = -72-4·3·4
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*3}=\frac{6}{6}
=1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*3}=\frac{8}{6}
=1+1/3
$