1/3k-2k+4=10

Solution for 1/3k-2k+4=10 equation:

1/3k-2k+4=10

We move all terms to the left:

1/3k-2k+4-(10)=0


Domain of the equation: 3k!=0

k!=0/3

k!=0

k∈R


We add all the numbers together, and all the variables

-2k+1/3k-6=0

We multiply all the terms by the denominator


-2k*3k-6*3k+1=0

Wy multiply elements

-6k^2-18k+1=0

a = -6; b = -18; c = +1;
Δ = b2-4ac
Δ = -182-4·(-6)·1
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{87}}{2*-6}=\frac{18-2\sqrt{87}}{-12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{87}}{2*-6}=\frac{18+2\sqrt{87}}{-12}
$