0=-16x^2+68

Solution for 0=-16x^2+68 equation:

0=-16x^2+68

We move all terms to the left:

0-(-16x^2+68)=0

We add all the numbers together, and all the variables

-(-16x^2+68)=0

We get rid of parentheses

16x^2-68=0

a = 16; b = 0; c = -68;
Δ = b2-4ac
Δ = 02-4·16·(-68)
Δ = 4352
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:

$\sqrt{\Delta}=\sqrt{4352}=\sqrt{256*17}=\sqrt{256}*\sqrt{17}=16\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{17}}{2*16}=\frac{0-16\sqrt{17}}{32}
=-\frac{16\sqrt{17}}{32}
=-\frac{\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{17}}{2*16}=\frac{0+16\sqrt{17}}{32}
=\frac{16\sqrt{17}}{32}
=\frac{\sqrt{17}}{2}
$