20x-x^2=16

Solution for 20x-x^2=16 equation:

20x-x^2=16

We move all terms to the left:

20x-x^2-(16)=0

We add all the numbers together, and all the variables

-1x^2+20x-16=0

a = -1; b = 20; c = -16;
Δ = b2-4ac
Δ = 202-4·(-1)·(-16)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{21}}{2*-1}=\frac{-20-4\sqrt{21}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{21}}{2*-1}=\frac{-20+4\sqrt{21}}{-2}
$