0=-16x^2+38

Solution for 0=-16x^2+38 equation:

0=-16x^2+38

We move all terms to the left:

0-(-16x^2+38)=0

We add all the numbers together, and all the variables

-(-16x^2+38)=0

We get rid of parentheses

16x^2-38=0

a = 16; b = 0; c = -38;
Δ = b2-4ac
Δ = 02-4·16·(-38)
Δ = 2432
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:

$\sqrt{\Delta}=\sqrt{2432}=\sqrt{64*38}=\sqrt{64}*\sqrt{38}=8\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{38}}{2*16}=\frac{0-8\sqrt{38}}{32}
=-\frac{8\sqrt{38}}{32}
=-\frac{\sqrt{38}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{38}}{2*16}=\frac{0+8\sqrt{38}}{32}
=\frac{8\sqrt{38}}{32}
=\frac{\sqrt{38}}{4}
$