2(10-2k)=-4k(-2k-2)

Solution for 2(10-2k)=-4k(-2k-2) equation:

2(10-2k)=-4k(-2k-2)

We move all terms to the left:

2(10-2k)-(-4k(-2k-2))=0

We add all the numbers together, and all the variables

2(-2k+10)-(-4k(-2k-2))=0

We multiply parentheses

-4k-(-4k(-2k-2))+20=0


We calculate terms in parentheses: -(-4k(-2k-2)), so:

-4k(-2k-2)

We multiply parentheses

8k^2+8k

Back to the equation:

-(8k^2+8k)


We get rid of parentheses

-8k^2-4k-8k+20=0

We add all the numbers together, and all the variables

-8k^2-12k+20=0

a = -8; b = -12; c = +20;
Δ = b2-4ac
Δ = -122-4·(-8)·20
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-28}{2*-8}=\frac{-16}{-16}
=1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+28}{2*-8}=\frac{40}{-16}
=-2+1/2
$