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13x^2-25x+6=0
a = 13; b = -25; c = +6;
Δ = b2-4ac
Δ = -252-4·13·6
Δ = 313
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{-(-25)-\sqrt{313}}{2*13}=\frac{25-\sqrt{313}}{26} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{313}}{2*13}=\frac{25+\sqrt{313}}{26} $
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