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x2-20x+1=0
We add all the numbers together, and all the variables
x^2-20x+1=0
a = 1; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·1·1
Δ = 396
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-6\sqrt{11}}{2*1}=\frac{20-6\sqrt{11}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+6\sqrt{11}}{2*1}=\frac{20+6\sqrt{11}}{2} $
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