15=3+180x+-16x^2

Solution for 15=3+180x+-16x^2 equation:

15=3+180x+-16x^2

We move all terms to the left:

15-(3+180x+-16x^2)=0

We use the square of the difference formula

-(3+180x-16x^2)+15=0

We get rid of parentheses

16x^2-180x-3+15=0

We add all the numbers together, and all the variables

16x^2-180x+12=0

a = 16; b = -180; c = +12;
Δ = b2-4ac
Δ = -1802-4·16·12
Δ = 31632
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:

$\sqrt{\Delta}=\sqrt{31632}=\sqrt{16*1977}=\sqrt{16}*\sqrt{1977}=4\sqrt{1977}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-180)-4\sqrt{1977}}{2*16}=\frac{180-4\sqrt{1977}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-180)+4\sqrt{1977}}{2*16}=\frac{180+4\sqrt{1977}}{32}
$