2z+1/4z=9

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

2z+1/4z=9

We move all terms to the left:

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


Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R


We multiply all the terms by the denominator


2z*4z-9*4z+1=0

Wy multiply elements

8z^2-36z+1=0

a = 8; b = -36; c = +1;
Δ = b2-4ac
Δ = -362-4·8·1
Δ = 1264
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:

$\sqrt{\Delta}=\sqrt{1264}=\sqrt{16*79}=\sqrt{16}*\sqrt{79}=4\sqrt{79}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{79}}{2*8}=\frac{36-4\sqrt{79}}{16}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{79}}{2*8}=\frac{36+4\sqrt{79}}{16}
$