(z-3)(z-5)=z

Solution for (z-3)(z-5)=z equation:

(z-3)(z-5)=z

We move all terms to the left:

(z-3)(z-5)-(z)=0

We add all the numbers together, and all the variables

-1z+(z-3)(z-5)=0

We multiply parentheses ..

(+z^2-5z-3z+15)-1z=0

We get rid of parentheses

z^2-5z-3z-1z+15=0

We add all the numbers together, and all the variables

z^2-9z+15=0

a = 1; b = -9; c = +15;
Δ = b2-4ac
Δ = -92-4·1·15
Δ = 21
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{-(-9)-\sqrt{21}}{2*1}=\frac{9-\sqrt{21}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{21}}{2*1}=\frac{9+\sqrt{21}}{2}
$