2-2(1-3k)^(1/2)=4k

Solution for 2-2(1-3k)^(1/2)=4k equation:

2-2(1-3k)^(1/2)=4k

We move all terms to the left:

2-2(1-3k)^(1/2)-(4k)=0

We add all the numbers together, and all the variables

-2(-3k+1)^(+1/2)-4k+2=0

We add all the numbers together, and all the variables

-4k-2(-3k+1)^(+1/2)+2=0

We multiply all the terms by the denominator


-4k*2)-2(-3k+1+1)^(+2*2)=0

We add all the numbers together, and all the variables

-4k*2)-2(-3k+2)^4=0

We add all the numbers together, and all the variables

-3k-4k*2)-2(=0

Wy multiply elements

-8k^2-3k=0

a = -8; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-8)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-8}=\frac{0}{-16}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-8}=\frac{6}{-16}
=-3/8
$