20x-x^2=40

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



20x-x^2=40
We move all terms to the left:
20x-x^2-(40)=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 = 240
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:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{15}}{2*-1}=\frac{-20-4\sqrt{15}}{-2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{15}}{2*-1}=\frac{-20+4\sqrt{15}}{-2} $

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