z/2z2+1=9

Solution for z/2z2+1=9 equation:

z/2z^2+1=9

We move all terms to the left:

z/2z^2+1-(9)=0


Domain of the equation: 2z^2!=0

z^2!=0/2

z^2!=√0

z!=0

z∈R


We add all the numbers together, and all the variables

z/2z^2-8=0

We multiply all the terms by the denominator


z-8*2z^2=0

Wy multiply elements

-16z^2+z=0

a = -16; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-16)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-16}=\frac{-2}{-32}
=1/16
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-16}=\frac{0}{-32}
=0
$