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x^2-20x-129=0
a = 1; b = -20; c = -129;
Δ = b2-4ac
Δ = -202-4·1·(-129)
Δ = 916
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{916}=\sqrt{4*229}=\sqrt{4}*\sqrt{229}=2\sqrt{229}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{229}}{2*1}=\frac{20-2\sqrt{229}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{229}}{2*1}=\frac{20+2\sqrt{229}}{2} $
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