2-3-4z=1/8z+9

Solution for 2-3-4z=1/8z+9 equation:

2-3-4z=1/8z+9

We move all terms to the left:

2-3-4z-(1/8z+9)=0


Domain of the equation: 8z+9)!=0

z∈R


We add all the numbers together, and all the variables

-4z-(1/8z+9)-1=0

We get rid of parentheses

-4z-1/8z-9-1=0

We multiply all the terms by the denominator


-4z*8z-9*8z-1*8z-1=0

Wy multiply elements

-32z^2-72z-8z-1=0

We add all the numbers together, and all the variables

-32z^2-80z-1=0

a = -32; b = -80; c = -1;
Δ = b2-4ac
Δ = -802-4·(-32)·(-1)
Δ = 6272
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:

$\sqrt{\Delta}=\sqrt{6272}=\sqrt{3136*2}=\sqrt{3136}*\sqrt{2}=56\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-56\sqrt{2}}{2*-32}=\frac{80-56\sqrt{2}}{-64}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+56\sqrt{2}}{2*-32}=\frac{80+56\sqrt{2}}{-64}
$