0=-16x^2+1362

Solution for 0=-16x^2+1362 equation:

0=-16x^2+1362

We move all terms to the left:

0-(-16x^2+1362)=0

We add all the numbers together, and all the variables

-(-16x^2+1362)=0

We get rid of parentheses

16x^2-1362=0

a = 16; b = 0; c = -1362;
Δ = b2-4ac
Δ = 02-4·16·(-1362)
Δ = 87168
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:

$\sqrt{\Delta}=\sqrt{87168}=\sqrt{64*1362}=\sqrt{64}*\sqrt{1362}=8\sqrt{1362}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{1362}}{2*16}=\frac{0-8\sqrt{1362}}{32}
=-\frac{8\sqrt{1362}}{32}
=-\frac{\sqrt{1362}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{1362}}{2*16}=\frac{0+8\sqrt{1362}}{32}
=\frac{8\sqrt{1362}}{32}
=\frac{\sqrt{1362}}{4}
$