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w^2-13w+8=0
a = 1; b = -13; c = +8;
Δ = b2-4ac
Δ = -132-4·1·8
Δ = 137
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{137}}{2*1}=\frac{13-\sqrt{137}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{137}}{2*1}=\frac{13+\sqrt{137}}{2} $
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