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4x^2-20x-2=0
a = 4; b = -20; c = -2;
Δ = b2-4ac
Δ = -202-4·4·(-2)
Δ = 432
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12\sqrt{3}}{2*4}=\frac{20-12\sqrt{3}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12\sqrt{3}}{2*4}=\frac{20+12\sqrt{3}}{8} $
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