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 = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{69}}{2*1}=\frac{-12-2\sqrt{69}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{69}}{2*1}=\frac{-12+2\sqrt{69}}{2}
$