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8x^2-26x+1=0
a = 8; b = -26; c = +1;
Δ = b2-4ac
Δ = -262-4·8·1
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{-(-26)-2\sqrt{161}}{2*8}=\frac{26-2\sqrt{161}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{161}}{2*8}=\frac{26+2\sqrt{161}}{16} $
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