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5x^2+13x+2=0
a = 5; b = 13; c = +2;
Δ = b2-4ac
Δ = 132-4·5·2
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*5}=\frac{-13-\sqrt{129}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{129}}{2*5}=\frac{-13+\sqrt{129}}{10} $
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