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10x^2-39x+33=0
a = 10; b = -39; c = +33;
Δ = b2-4ac
Δ = -392-4·10·33
Δ = 201
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{-(-39)-\sqrt{201}}{2*10}=\frac{39-\sqrt{201}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{201}}{2*10}=\frac{39+\sqrt{201}}{20} $
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