500=3+180x-16x^2

Solution for 500=3+180x-16x^2 equation:

500=3+180x-16x^2

We move all terms to the left:

500-(3+180x-16x^2)=0

We get rid of parentheses

16x^2-180x-3+500=0

We add all the numbers together, and all the variables

16x^2-180x+497=0

a = 16; b = -180; c = +497;
Δ = b2-4ac
Δ = -1802-4·16·497
Δ = 592
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:

$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-180)-4\sqrt{37}}{2*16}=\frac{180-4\sqrt{37}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-180)+4\sqrt{37}}{2*16}=\frac{180+4\sqrt{37}}{32}
$