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9x^2-40x+1=0
a = 9; b = -40; c = +1;
Δ = b2-4ac
Δ = -402-4·9·1
Δ = 1564
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1564}=\sqrt{4*391}=\sqrt{4}*\sqrt{391}=2\sqrt{391}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{391}}{2*9}=\frac{40-2\sqrt{391}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{391}}{2*9}=\frac{40+2\sqrt{391}}{18} $
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