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

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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 $

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