0=-192+36x^2

Solution for 0=-192+36x^2 equation:

0=-192+36x^2

We move all terms to the left:

0-(-192+36x^2)=0

We add all the numbers together, and all the variables

-(-192+36x^2)=0

We get rid of parentheses

-36x^2+192=0

a = -36; b = 0; c = +192;
Δ = b2-4ac
Δ = 02-4·(-36)·192
Δ = 27648
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:

$\sqrt{\Delta}=\sqrt{27648}=\sqrt{9216*3}=\sqrt{9216}*\sqrt{3}=96\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-96\sqrt{3}}{2*-36}=\frac{0-96\sqrt{3}}{-72}
=-\frac{96\sqrt{3}}{-72}
=-\frac{4\sqrt{3}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+96\sqrt{3}}{2*-36}=\frac{0+96\sqrt{3}}{-72}
=\frac{96\sqrt{3}}{-72}
=\frac{4\sqrt{3}}{-3}
$