k^2+9=-10k

Solution for k^2+9=-10k equation:

k^2+9=-10k

We move all terms to the left:

k^2+9-(-10k)=0

We get rid of parentheses

k^2+10k+9=0

a = 1; b = 10; c = +9;
Δ = b2-4ac
Δ = 102-4·1·9
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8}{2*1}=\frac{-18}{2}
=-9
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8}{2*1}=\frac{-2}{2}
=-1
$