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

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
$