(k-4)(k-1)=18

Solution for (k-4)(k-1)=18 equation:

(k-4)(k-1)=18

We move all terms to the left:

(k-4)(k-1)-(18)=0

We multiply parentheses ..

(+k^2-1k-4k+4)-18=0

We get rid of parentheses

k^2-1k-4k+4-18=0

We add all the numbers together, and all the variables

k^2-5k-14=0

a = 1; b = -5; c = -14;
Δ = b2-4ac
Δ = -52-4·1·(-14)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*1}=\frac{-4}{2}
=-2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*1}=\frac{14}{2}
=7
$