100=0.22xD^2

Solution for 100=0.22xD^2 equation:

100=0.22x^2

We move all terms to the left:

100-(0.22x^2)=0

We get rid of parentheses

-0.22x^2+100=0

a = -0.22; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-0.22)·100
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{22}}{2*-0.22}=\frac{0-2\sqrt{22}}{-0.44}
=-\frac{2\sqrt{22}}{-0.44}
=-\frac{\sqrt{22}}{-0.22}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{22}}{2*-0.22}=\frac{0+2\sqrt{22}}{-0.44}
=\frac{2\sqrt{22}}{-0.44}
=\frac{\sqrt{22}}{-0.22}
$