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x^2-41x+8=0
a = 1; b = -41; c = +8;
Δ = b2-4ac
Δ = -412-4·1·8
Δ = 1649
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-\sqrt{1649}}{2*1}=\frac{41-\sqrt{1649}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+\sqrt{1649}}{2*1}=\frac{41+\sqrt{1649}}{2} $
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