1k^2-16=-10

Solution for 1k^2-16=-10 equation:

1k^2-16=-10

We move all terms to the left:

1k^2-16-(-10)=0

We add all the numbers together, and all the variables

k^2-6=0

a = 1; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·1·(-6)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*1}=\frac{0-2\sqrt{6}}{2}
=-\frac{2\sqrt{6}}{2}
=-\sqrt{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*1}=\frac{0+2\sqrt{6}}{2}
=\frac{2\sqrt{6}}{2}
=\sqrt{6}
$