168=20+1k(k+20)

Solution for 168=20+1k(k+20) equation:

168=20+1k(k+20)

We move all terms to the left:

168-(20+1k(k+20))=0


We calculate terms in parentheses: -(20+1k(k+20)), so:

20+1k(k+20)

determiningTheFunctionDomain
1k(k+20)+20

We multiply parentheses

k^2+20k+20

Back to the equation:

-(k^2+20k+20)


We get rid of parentheses

-k^2-20k-20+168=0

We add all the numbers together, and all the variables

-1k^2-20k+148=0

a = -1; b = -20; c = +148;
Δ = b2-4ac
Δ = -202-4·(-1)·148
Δ = 992
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{62}}{2*-1}=\frac{20-4\sqrt{62}}{-2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{62}}{2*-1}=\frac{20+4\sqrt{62}}{-2}
$