(11/5)z=2/5

Solution for (11/5)z=2/5 equation:

(11/5)z=2/5

We move all terms to the left:

(11/5)z-(2/5)=0


Domain of the equation: 5)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+11/5)z-(+2/5)=0

We multiply parentheses

11z^2-(+2/5)=0

We get rid of parentheses

11z^2-2/5=0

We multiply all the terms by the denominator


11z^2*5-2=0

Wy multiply elements

55z^2-2=0

a = 55; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·55·(-2)
Δ = 440
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{110}}{2*55}=\frac{0-2\sqrt{110}}{110}
=-\frac{2\sqrt{110}}{110}
=-\frac{\sqrt{110}}{55}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{110}}{2*55}=\frac{0+2\sqrt{110}}{110}
=\frac{2\sqrt{110}}{110}
=\frac{\sqrt{110}}{55}
$