(z^2-z)/2=2z-3

Solution for (z^2-z)/2=2z-3 equation:

(z^2-z)/2=2z-3

We move all terms to the left:

(z^2-z)/2-(2z-3)=0

We get rid of parentheses

(z^2-z)/2-2z+3=0

We multiply all the terms by the denominator


(z^2-z)-2z*2+3*2=0

We add all the numbers together, and all the variables

(z^2-z)-2z*2+6=0

Wy multiply elements

(z^2-z)-4z+6=0

We get rid of parentheses

z^2-z-4z+6=0

We add all the numbers together, and all the variables

z^2-5z+6=0

a = 1; b = -5; c = +6;
Δ = b2-4ac
Δ = -52-4·1·6
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{-(-5)-1}{2*1}=\frac{4}{2}
=2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*1}=\frac{6}{2}
=3
$