z^2=12z+36=0

Solution for z^2=12z+36=0 equation:

z^2=12z+36=0

We move all terms to the left:

z^2-(12z+36)=0

We get rid of parentheses

z^2-12z-36=0

a = 1; b = -12; c = -36;
Δ = b2-4ac
Δ = -122-4·1·(-36)
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{2}}{2*1}=\frac{12-12\sqrt{2}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{2}}{2*1}=\frac{12+12\sqrt{2}}{2}
$