0=-16x^2+1053

Solution for 0=-16x^2+1053 equation:

0=-16x^2+1053

We move all terms to the left:

0-(-16x^2+1053)=0

We add all the numbers together, and all the variables

-(-16x^2+1053)=0

We get rid of parentheses

16x^2-1053=0

a = 16; b = 0; c = -1053;
Δ = b2-4ac
Δ = 02-4·16·(-1053)
Δ = 67392
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:

$\sqrt{\Delta}=\sqrt{67392}=\sqrt{5184*13}=\sqrt{5184}*\sqrt{13}=72\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-72\sqrt{13}}{2*16}=\frac{0-72\sqrt{13}}{32}
=-\frac{72\sqrt{13}}{32}
=-\frac{9\sqrt{13}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+72\sqrt{13}}{2*16}=\frac{0+72\sqrt{13}}{32}
=\frac{72\sqrt{13}}{32}
=\frac{9\sqrt{13}}{4}
$