k^2+12k+33=0

Solution for k^2+12k+33=0 equation:

k^2+12k+33=0

a = 1; b = 12; c = +33;
Δ = b2-4ac
Δ = 122-4·1·33
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{3}}{2*1}=\frac{-12-2\sqrt{3}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{3}}{2*1}=\frac{-12+2\sqrt{3}}{2}
$