3/4z-4=-7/8z+1

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

3/4z-4=-7/8z+1

We move all terms to the left:

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


Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R



Domain of the equation: 8z+1)!=0

z∈R


We get rid of parentheses

3/4z+7/8z-1-4=0

We calculate fractions


24z/32z^2+28z/32z^2-1-4=0

We add all the numbers together, and all the variables

24z/32z^2+28z/32z^2-5=0

We multiply all the terms by the denominator


24z+28z-5*32z^2=0

We add all the numbers together, and all the variables

52z-5*32z^2=0

Wy multiply elements

-160z^2+52z=0

a = -160; b = 52; c = 0;
Δ = b2-4ac
Δ = 522-4·(-160)·0
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-52}{2*-160}=\frac{-104}{-320}
=13/40
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+52}{2*-160}=\frac{0}{-320}
=0
$