1/5z+8/10z+3=7

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Solution for 1/5z+8/10z+3=7 equation:



1/5z+8/10z+3=7
We move all terms to the left:
1/5z+8/10z+3-(7)=0
Domain of the equation: 5z!=0
z!=0/5
z!=0
z∈R
Domain of the equation: 10z!=0
z!=0/10
z!=0
z∈R
We add all the numbers together, and all the variables
1/5z+8/10z-4=0
We calculate fractions
10z/50z^2+40z/50z^2-4=0
We multiply all the terms by the denominator
10z+40z-4*50z^2=0
We add all the numbers together, and all the variables
50z-4*50z^2=0
Wy multiply elements
-200z^2+50z=0
a = -200; b = 50; c = 0;
Δ = b2-4ac
Δ = 502-4·(-200)·0
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-50}{2*-200}=\frac{-100}{-400} =1/4 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+50}{2*-200}=\frac{0}{-400} =0 $

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