0=-16x^2+540

Solution for 0=-16x^2+540 equation:

0=-16x^2+540

We move all terms to the left:

0-(-16x^2+540)=0

We add all the numbers together, and all the variables

-(-16x^2+540)=0

We get rid of parentheses

16x^2-540=0

a = 16; b = 0; c = -540;
Δ = b2-4ac
Δ = 02-4·16·(-540)
Δ = 34560
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:

$\sqrt{\Delta}=\sqrt{34560}=\sqrt{2304*15}=\sqrt{2304}*\sqrt{15}=48\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{15}}{2*16}=\frac{0-48\sqrt{15}}{32}
=-\frac{48\sqrt{15}}{32}
=-\frac{3\sqrt{15}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{15}}{2*16}=\frac{0+48\sqrt{15}}{32}
=\frac{48\sqrt{15}}{32}
=\frac{3\sqrt{15}}{2}
$