1+-3/4k=-7/8k+2

Solution for 1+-3/4k=-7/8k+2 equation:

1+-3/4k=-7/8k+2

We move all terms to the left:

1+-3/4k-(-7/8k+2)=0


Domain of the equation: 4k!=0

k!=0/4

k!=0

k∈R



Domain of the equation: 8k+2)!=0

k∈R


We add all the numbers together, and all the variables

-3/4k-(-7/8k+2)=0

We get rid of parentheses

-3/4k+7/8k-2=0

We calculate fractions


(-24k)/32k^2+28k/32k^2-2=0

We multiply all the terms by the denominator


(-24k)+28k-2*32k^2=0

We add all the numbers together, and all the variables

28k+(-24k)-2*32k^2=0

Wy multiply elements

-64k^2+28k+(-24k)=0

We get rid of parentheses

-64k^2+28k-24k=0

We add all the numbers together, and all the variables

-64k^2+4k=0

a = -64; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-64)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-64}=\frac{-8}{-128}
=1/16
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-64}=\frac{0}{-128}
=0
$