k2+12k+28=-7

Solution for k2+12k+28=-7 equation:

k2+12k+28=-7

We move all terms to the left:

k2+12k+28-(-7)=0

We add all the numbers together, and all the variables

k^2+12k+35=0

a = 1; b = 12; c = +35;
Δ = b2-4ac
Δ = 122-4·1·35
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2}{2*1}=\frac{-14}{2}
=-7
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2}{2*1}=\frac{-10}{2}
=-5
$