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4X^2-20X+5=0
a = 4; b = -20; c = +5;
Δ = b2-4ac
Δ = -202-4·4·5
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*4}=\frac{20-8\sqrt{5}}{8} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{5}}{2*4}=\frac{20+8\sqrt{5}}{8} $
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