1=1.04k^2

Solution for 1=1.04k^2 equation:

1=1.04k^2

We move all terms to the left:

1-(1.04k^2)=0

We get rid of parentheses

-1.04k^2+1=0

a = -1.04; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-1.04)·1
Δ = 4.16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{4.16}}{2*-1.04}=\frac{0-\sqrt{4.16}}{-2.08}
=-\frac{\sqrt{}}{-2.08}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{4.16}}{2*-1.04}=\frac{0+\sqrt{4.16}}{-2.08}
=\frac{\sqrt{}}{-2.08}
$