-3z+5=z^2

Solution for -3z+5=z^2 equation:

-3z+5=z^2

We move all terms to the left:

-3z+5-(z^2)=0

determiningTheFunctionDomain
-z^2-3z+5=0

We add all the numbers together, and all the variables

-1z^2-3z+5=0

a = -1; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-1)·5
Δ = 29
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{29}}{2*-1}=\frac{3-\sqrt{29}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{29}}{2*-1}=\frac{3+\sqrt{29}}{-2}
$