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13x^2-29x+3=0
a = 13; b = -29; c = +3;
Δ = b2-4ac
Δ = -292-4·13·3
Δ = 685
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{-(-29)-\sqrt{685}}{2*13}=\frac{29-\sqrt{685}}{26} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{685}}{2*13}=\frac{29+\sqrt{685}}{26} $
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