4/5z-3=9/10z+2

Solution for 4/5z-3=9/10z+2 equation:

4/5z-3=9/10z+2

We move all terms to the left:

4/5z-3-(9/10z+2)=0


Domain of the equation: 5z!=0

z!=0/5

z!=0

z∈R



Domain of the equation: 10z+2)!=0

z∈R


We get rid of parentheses

4/5z-9/10z-2-3=0

We calculate fractions


40z/50z^2+(-45z)/50z^2-2-3=0

We add all the numbers together, and all the variables

40z/50z^2+(-45z)/50z^2-5=0

We multiply all the terms by the denominator


40z+(-45z)-5*50z^2=0

Wy multiply elements

-250z^2+40z+(-45z)=0

We get rid of parentheses

-250z^2+40z-45z=0

We add all the numbers together, and all the variables

-250z^2-5z=0

a = -250; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-250)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-250}=\frac{0}{-500}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-250}=\frac{10}{-500}
=-1/50
$