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

Solution for k^2(k-10)(k+10)=0 equation:

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

We use the square of the difference formula

k^2-100=0

a = 1; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·1·(-100)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*1}=\frac{-20}{2}
=-10
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*1}=\frac{20}{2}
=10
$