z=2,3z^2

Solution for z=2,3z^2 equation:

z=2.3z^2

We move all terms to the left:

z-(2.3z^2)=0

We get rid of parentheses

-2.3z^2+z=0

a = -2.3; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-2.3)·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*-2.3}=\frac{-2}{-4.6}
=1/2.3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-2.3}=\frac{0}{-4.6}
=0
$