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3x^2-39x+125=0
a = 3; b = -39; c = +125;
Δ = b2-4ac
Δ = -392-4·3·125
Δ = 21
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{21}}{2*3}=\frac{39-\sqrt{21}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{21}}{2*3}=\frac{39+\sqrt{21}}{6} $
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