2(k-1)=10k(k-2)

Solution for 2(k-1)=10k(k-2) equation:

2(k-1)=10k(k-2)

We move all terms to the left:

2(k-1)-(10k(k-2))=0

We multiply parentheses

2k-(10k(k-2))-2=0


We calculate terms in parentheses: -(10k(k-2)), so:

10k(k-2)

We multiply parentheses

10k^2-20k

Back to the equation:

-(10k^2-20k)


We get rid of parentheses

-10k^2+2k+20k-2=0

We add all the numbers together, and all the variables

-10k^2+22k-2=0

a = -10; b = 22; c = -2;
Δ = b2-4ac
Δ = 222-4·(-10)·(-2)
Δ = 404
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:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{101}}{2*-10}=\frac{-22-2\sqrt{101}}{-20}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{101}}{2*-10}=\frac{-22+2\sqrt{101}}{-20}
$