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

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

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

We move all terms to the left:

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


Domain of the equation: 7k!=0

k!=0/7

k!=0

k∈R



Domain of the equation: 4k+4)!=0

k∈R


We get rid of parentheses

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

We calculate fractions


4k/28k^2+(-21k)/28k^2-4-6=0

We add all the numbers together, and all the variables

4k/28k^2+(-21k)/28k^2-10=0

We multiply all the terms by the denominator


4k+(-21k)-10*28k^2=0

Wy multiply elements

-280k^2+4k+(-21k)=0

We get rid of parentheses

-280k^2+4k-21k=0

We add all the numbers together, and all the variables

-280k^2-17k=0

a = -280; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·(-280)·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-17}{2*-280}=\frac{0}{-560}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*-280}=\frac{34}{-560}
=-17/280
$