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20x^2-2x-8=0
a = 20; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·20·(-8)
Δ = 644
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{161}}{2*20}=\frac{2-2\sqrt{161}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{161}}{2*20}=\frac{2+2\sqrt{161}}{40} $
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