z^2=-41-17z

Solution for z^2=-41-17z equation:

z^2=-41-17z

We move all terms to the left:

z^2-(-41-17z)=0

We add all the numbers together, and all the variables

z^2-(-17z-41)=0

We get rid of parentheses

z^2+17z+41=0

a = 1; b = 17; c = +41;
Δ = b2-4ac
Δ = 172-4·1·41
Δ = 125
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:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-5\sqrt{5}}{2*1}=\frac{-17-5\sqrt{5}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+5\sqrt{5}}{2*1}=\frac{-17+5\sqrt{5}}{2}
$