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8x^2-41x+21=0
a = 8; b = -41; c = +21;
Δ = b2-4ac
Δ = -412-4·8·21
Δ = 1009
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{1009}}{2*8}=\frac{41-\sqrt{1009}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+\sqrt{1009}}{2*8}=\frac{41+\sqrt{1009}}{16} $
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