1/5z-3/4z=2

Solution for 1/5z-3/4z=2 equation:

1/5z-3/4z=2

We move all terms to the left:

1/5z-3/4z-(2)=0


Domain of the equation: 5z!=0

z!=0/5

z!=0

z∈R



Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R


We calculate fractions


4z/20z^2+(-15z)/20z^2-2=0

We multiply all the terms by the denominator


4z+(-15z)-2*20z^2=0

Wy multiply elements

-40z^2+4z+(-15z)=0

We get rid of parentheses

-40z^2+4z-15z=0

We add all the numbers together, and all the variables

-40z^2-11z=0

a = -40; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·(-40)·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*-40}=\frac{0}{-80}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*-40}=\frac{22}{-80}
=-11/40
$