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10x^2+13x-4=0
a = 10; b = 13; c = -4;
Δ = b2-4ac
Δ = 132-4·10·(-4)
Δ = 329
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{329}}{2*10}=\frac{-13-\sqrt{329}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{329}}{2*10}=\frac{-13+\sqrt{329}}{20} $
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