-(4/9)z=-8

Solution for -(4/9)z=-8 equation:

-(4/9)z=-8

We move all terms to the left:

-(4/9)z-(-8)=0


Domain of the equation: 9)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

-(+4/9)z-(-8)=0

We add all the numbers together, and all the variables

-(+4/9)z+8=0

We multiply parentheses

-4z^2+8=0

a = -4; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-4)·8
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*-4}=\frac{0-8\sqrt{2}}{-8}
=-\frac{8\sqrt{2}}{-8}
=-\frac{\sqrt{2}}{-1}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*-4}=\frac{0+8\sqrt{2}}{-8}
=\frac{8\sqrt{2}}{-8}
=\frac{\sqrt{2}}{-1}
$