-3z+1=-z2

Solution for -3z+1=-z2 equation:

-3z+1=-z2

We move all terms to the left:

-3z+1-(-z2)=0

We add all the numbers together, and all the variables

-(-1z^2)-3z+1=0

We get rid of parentheses

1z^2-3z+1=0

We add all the numbers together, and all the variables

z^2-3z+1=0

a = 1; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·1·1
Δ = 5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{5}}{2*1}=\frac{3-\sqrt{5}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{5}}{2*1}=\frac{3+\sqrt{5}}{2}
$