6/5k-1/4=3+7/6k

Solution for 6/5k-1/4=3+7/6k equation:

6/5k-1/4=3+7/6k

We move all terms to the left:

6/5k-1/4-(3+7/6k)=0


Domain of the equation: 5k!=0

k!=0/5

k!=0

k∈R



Domain of the equation: 6k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

6/5k-(7/6k+3)-1/4=0

We get rid of parentheses

6/5k-7/6k-3-1/4=0

We calculate fractions


(-180k^2)/480k^2+576k/480k^2+(-560k)/480k^2-3=0

We multiply all the terms by the denominator


(-180k^2)+576k+(-560k)-3*480k^2=0

Wy multiply elements

(-180k^2)-1440k^2+576k+(-560k)=0

We get rid of parentheses

-180k^2-1440k^2+576k-560k=0

We add all the numbers together, and all the variables

-1620k^2+16k=0

a = -1620; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-1620)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-1620}=\frac{-32}{-3240}
=4/405
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-1620}=\frac{0}{-3240}
=0
$