40-20x-x^2=0

Solution for 40-20x-x^2=0 equation:

40-20x-x^2=0

We add all the numbers together, and all the variables

-1x^2-20x+40=0

a = -1; b = -20; c = +40;
Δ = b2-4ac
Δ = -202-4·(-1)·40
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{35}}{2*-1}=\frac{20-4\sqrt{35}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{35}}{2*-1}=\frac{20+4\sqrt{35}}{-2}
$