x^2=2x+447=12

Solution for x^2=2x+447=12 equation:

x^2=2x+447=12

We move all terms to the left:

x^2-(2x+447)=0

We get rid of parentheses

x^2-2x-447=0

a = 1; b = -2; c = -447;
Δ = b2-4ac
Δ = -22-4·1·(-447)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-16\sqrt{7}}{2*1}=\frac{2-16\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+16\sqrt{7}}{2*1}=\frac{2+16\sqrt{7}}{2}
$