z^2+12z=-13

Solution for z^2+12z=-13 equation:

z^2+12z=-13

We move all terms to the left:

z^2+12z-(-13)=0

We add all the numbers together, and all the variables

z^2+12z+13=0

a = 1; b = 12; c = +13;
Δ = b2-4ac
Δ = 122-4·1·13
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{23}}{2*1}=\frac{-12-2\sqrt{23}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{23}}{2*1}=\frac{-12+2\sqrt{23}}{2}
$