500+12x-x^2=0

Solution for 500+12x-x^2=0 equation:

500+12x-x^2=0

We add all the numbers together, and all the variables

-1x^2+12x+500=0

a = -1; b = 12; c = +500;
Δ = b2-4ac
Δ = 122-4·(-1)·500
Δ = 2144
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:

$\sqrt{\Delta}=\sqrt{2144}=\sqrt{16*134}=\sqrt{16}*\sqrt{134}=4\sqrt{134}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{134}}{2*-1}=\frac{-12-4\sqrt{134}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{134}}{2*-1}=\frac{-12+4\sqrt{134}}{-2}
$