3x^2+2x-2^32=0

Solution for 3x^2+2x-2^32=0 equation:

3x^2+2x-2^32=0

We add all the numbers together, and all the variables

3x^2+2x-4294967296=0

a = 3; b = 2; c = -4294967296;
Δ = b2-4ac
Δ = 22-4·3·(-4294967296)
Δ = 51539607556
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:

$\sqrt{\Delta}=\sqrt{51539607556}=\sqrt{1444*35692249}=\sqrt{1444}*\sqrt{35692249}=38\sqrt{35692249}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-38\sqrt{35692249}}{2*3}=\frac{-2-38\sqrt{35692249}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+38\sqrt{35692249}}{2*3}=\frac{-2+38\sqrt{35692249}}{6}
$