(-4k+16k^2-4(-2k^2+11k-4)=-4

Solution for (-4k+16k^2-4(-2k^2+11k-4)=-4 equation:

(-4k+16k^2-4(-2k^2+11k-4)=-4

We move all terms to the left:

(-4k+16k^2-4(-2k^2+11k-4)-(-4)=0


We calculate terms in parentheses: +(-4k+16k^2-4(-2k^2+11k-4)-(-4), so:

-4k+16k^2-4(-2k^2+11k-4)-(-4

determiningTheFunctionDomain
16k^2-4(-2k^2+11k-4)-4k-(-4

We add all the numbers together, and all the variables

16k^2-4(-2k^2+11k-4)-4k

We multiply parentheses

16k^2+8k^2-44k-4k+16

We add all the numbers together, and all the variables

24k^2-48k+16

Back to the equation:

+(24k^2-48k+16)


We get rid of parentheses

24k^2-48k+16=0

a = 24; b = -48; c = +16;
Δ = b2-4ac
Δ = -482-4·24·16
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{3}}{2*24}=\frac{48-16\sqrt{3}}{48}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{3}}{2*24}=\frac{48+16\sqrt{3}}{48}
$