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4r^2-8r-1=0
a = 4; b = -8; c = -1;
Δ = b2-4ac
Δ = -82-4·4·(-1)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{5}}{2*4}=\frac{8-4\sqrt{5}}{8} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{5}}{2*4}=\frac{8+4\sqrt{5}}{8} $
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