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2x^2-13x+5=0
a = 2; b = -13; c = +5;
Δ = b2-4ac
Δ = -132-4·2·5
Δ = 129
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{129}}{2*2}=\frac{13-\sqrt{129}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{129}}{2*2}=\frac{13+\sqrt{129}}{4} $
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