z^2+4z=22=0

Solution for z^2+4z=22=0 equation:

z^2+4z=22=0

We move all terms to the left:

z^2+4z-(22)=0

a = 1; b = 4; c = -22;
Δ = b2-4ac
Δ = 42-4·1·(-22)
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{26}}{2*1}=\frac{-4-2\sqrt{26}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{26}}{2*1}=\frac{-4+2\sqrt{26}}{2}
$