7/4k+1/5k-3/2=4/5k

Solution for 7/4k+1/5k-3/2=4/5k equation:

7/4k+1/5k-3/2=4/5k

We move all terms to the left:

7/4k+1/5k-3/2-(4/5k)=0


Domain of the equation: 4k!=0

k!=0/4

k!=0

k∈R



Domain of the equation: 5k!=0

k!=0/5

k!=0

k∈R



Domain of the equation: 5k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

7/4k+1/5k-(+4/5k)-3/2=0

We get rid of parentheses

7/4k+1/5k-4/5k-3/2=0

We calculate fractions


(-300k^2)/80k^2+140k/80k^2+(-64k+1)/80k^2=0

We multiply all the terms by the denominator


(-300k^2)+140k+(-64k+1)=0

We get rid of parentheses

-300k^2+140k-64k+1=0

We add all the numbers together, and all the variables

-300k^2+76k+1=0

a = -300; b = 76; c = +1;
Δ = b2-4ac
Δ = 762-4·(-300)·1
Δ = 6976
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:

$\sqrt{\Delta}=\sqrt{6976}=\sqrt{64*109}=\sqrt{64}*\sqrt{109}=8\sqrt{109}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(76)-8\sqrt{109}}{2*-300}=\frac{-76-8\sqrt{109}}{-600}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(76)+8\sqrt{109}}{2*-300}=\frac{-76+8\sqrt{109}}{-600}
$