(K+9)^2-4(k^2+15)=0

Solution for (K+9)^2-4(k^2+15)=0 equation:

(+9)^2-4(K^2+15)=0

We add all the numbers together, and all the variables

-4(K^2+15)+9^2=0

We add all the numbers together, and all the variables

-4(K^2+15)+81=0

We multiply parentheses

-4K^2-60+81=0

We add all the numbers together, and all the variables

-4K^2+21=0

a = -4; b = 0; c = +21;
Δ = b2-4ac
Δ = 02-4·(-4)·21
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{21}}{2*-4}=\frac{0-4\sqrt{21}}{-8}
=-\frac{4\sqrt{21}}{-8}
=-\frac{\sqrt{21}}{-2}
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{21}}{2*-4}=\frac{0+4\sqrt{21}}{-8}
=\frac{4\sqrt{21}}{-8}
=\frac{\sqrt{21}}{-2}
$