k^2+18k+21=0

Solution for k^2+18k+21=0 equation:

k^2+18k+21=0

a = 1; b = 18; c = +21;
Δ = b2-4ac
Δ = 182-4·1·21
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-4\sqrt{15}}{2*1}=\frac{-18-4\sqrt{15}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+4\sqrt{15}}{2*1}=\frac{-18+4\sqrt{15}}{2}
$