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2b^2-18b+1=0
a = 2; b = -18; c = +1;
Δ = b2-4ac
Δ = -182-4·2·1
Δ = 316
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{79}}{2*2}=\frac{18-2\sqrt{79}}{4} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{79}}{2*2}=\frac{18+2\sqrt{79}}{4} $
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