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2x^2-39x+2=0
a = 2; b = -39; c = +2;
Δ = b2-4ac
Δ = -392-4·2·2
Δ = 1505
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{1505}}{2*2}=\frac{39-\sqrt{1505}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1505}}{2*2}=\frac{39+\sqrt{1505}}{4} $
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