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a^2-20a+5=0
a = 1; b = -20; c = +5;
Δ = b2-4ac
Δ = -202-4·1·5
Δ = 380
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{380}=\sqrt{4*95}=\sqrt{4}*\sqrt{95}=2\sqrt{95}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{95}}{2*1}=\frac{20-2\sqrt{95}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{95}}{2*1}=\frac{20+2\sqrt{95}}{2} $
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