k^2+9k+2.25=0

Solution for k^2+9k+2.25=0 equation:

k^2+9k+2.25=0

a = 1; b = 9; c = +2.25;
Δ = b2-4ac
Δ = 92-4·1·2.25
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-6\sqrt{2}}{2*1}=\frac{-9-6\sqrt{2}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+6\sqrt{2}}{2*1}=\frac{-9+6\sqrt{2}}{2}
$