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11x^2-13x+3=0
a = 11; b = -13; c = +3;
Δ = b2-4ac
Δ = -132-4·11·3
Δ = 37
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{-(-13)-\sqrt{37}}{2*11}=\frac{13-\sqrt{37}}{22} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{37}}{2*11}=\frac{13+\sqrt{37}}{22} $
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