k^2+18K-63=0

Solution for k^2+18K-63=0 equation:

k^2+18-63=0

We add all the numbers together, and all the variables

k^2-45=0

a = 1; b = 0; c = -45;
Δ = b2-4ac
Δ = 02-4·1·(-45)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{5}}{2*1}=\frac{0-6\sqrt{5}}{2}
=-\frac{6\sqrt{5}}{2}
=-3\sqrt{5}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{5}}{2*1}=\frac{0+6\sqrt{5}}{2}
=\frac{6\sqrt{5}}{2}
=3\sqrt{5}
$