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x^2-20x=-20
We move all terms to the left:
x^2-20x-(-20)=0
We add all the numbers together, and all the variables
x^2-20x+20=0
a = 1; b = -20; c = +20;
Δ = b2-4ac
Δ = -202-4·1·20
Δ = 320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{5}}{2*1}=\frac{20-8\sqrt{5}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{5}}{2*1}=\frac{20+8\sqrt{5}}{2} $
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