z^2+26z=-49

Solution for z^2+26z=-49 equation:

z^2+26z=-49

We move all terms to the left:

z^2+26z-(-49)=0

We add all the numbers together, and all the variables

z^2+26z+49=0

a = 1; b = 26; c = +49;
Δ = b2-4ac
Δ = 262-4·1·49
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-4\sqrt{30}}{2*1}=\frac{-26-4\sqrt{30}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+4\sqrt{30}}{2*1}=\frac{-26+4\sqrt{30}}{2}
$