k^2+16k=-12

Solution for k^2+16k=-12 equation:

k^2+16k=-12

We move all terms to the left:

k^2+16k-(-12)=0

We add all the numbers together, and all the variables

k^2+16k+12=0

a = 1; b = 16; c = +12;
Δ = b2-4ac
Δ = 162-4·1·12
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{13}}{2*1}=\frac{-16-4\sqrt{13}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{13}}{2*1}=\frac{-16+4\sqrt{13}}{2}
$