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2w^2-13w+1=0
a = 2; b = -13; c = +1;
Δ = b2-4ac
Δ = -132-4·2·1
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{161}}{2*2}=\frac{13-\sqrt{161}}{4} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{161}}{2*2}=\frac{13+\sqrt{161}}{4} $
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