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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3k^2-11k-4=0

a = 3; b = -11; c = -4;
Δ = b2-4ac
Δ = -112-4·3·(-4)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*3}=\frac{-2}{6}
=-1/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*3}=\frac{24}{6}
=4
$