4z2+z=9z

Solution for 4z2+z=9z equation:

4z^2+z=9z

We move all terms to the left:

4z^2+z-(9z)=0

We add all the numbers together, and all the variables

4z^2-8z=0

a = 4; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·4·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*4}=\frac{0}{8}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*4}=\frac{16}{8}
=2
$