31-20x+3x^2=22483

Solution for 31-20x+3x^2=22483 equation:

31-20x+3x^2=22483

We move all terms to the left:

31-20x+3x^2-(22483)=0

We add all the numbers together, and all the variables

3x^2-20x-22452=0

a = 3; b = -20; c = -22452;
Δ = b2-4ac
Δ = -202-4·3·(-22452)
Δ = 269824
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:

$\sqrt{\Delta}=\sqrt{269824}=\sqrt{256*1054}=\sqrt{256}*\sqrt{1054}=16\sqrt{1054}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16\sqrt{1054}}{2*3}=\frac{20-16\sqrt{1054}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16\sqrt{1054}}{2*3}=\frac{20+16\sqrt{1054}}{6}
$