k^2+12k=28

Solution for k^2+12k=28 equation:

k^2+12k=28

We move all terms to the left:

k^2+12k-(28)=0

a = 1; b = 12; c = -28;
Δ = b2-4ac
Δ = 122-4·1·(-28)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16}{2*1}=\frac{-28}{2}
=-14
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16}{2*1}=\frac{4}{2}
=2
$