1/4k-4=3/7k+6

Solution for 1/4k-4=3/7k+6 equation:

1/4k-4=3/7k+6

We move all terms to the left:

1/4k-4-(3/7k+6)=0


Domain of the equation: 4k!=0

k!=0/4

k!=0

k∈R



Domain of the equation: 7k+6)!=0

k∈R


We get rid of parentheses

1/4k-3/7k-6-4=0

We calculate fractions


7k/28k^2+(-12k)/28k^2-6-4=0

We add all the numbers together, and all the variables

7k/28k^2+(-12k)/28k^2-10=0

We multiply all the terms by the denominator


7k+(-12k)-10*28k^2=0

Wy multiply elements

-280k^2+7k+(-12k)=0

We get rid of parentheses

-280k^2+7k-12k=0

We add all the numbers together, and all the variables

-280k^2-5k=0

a = -280; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-280)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-280}=\frac{0}{-560}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-280}=\frac{10}{-560}
=-1/56
$