z2-6=3z

Solution for z2-6=3z equation:

z2-6=3z

We move all terms to the left:

z2-6-(3z)=0

We add all the numbers together, and all the variables

z^2-3z-6=0

a = 1; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·1·(-6)
Δ = 33
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{33}}{2*1}=\frac{3-\sqrt{33}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*1}=\frac{3+\sqrt{33}}{2}
$