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2019x^2+x-2019=0
a = 2019; b = 1; c = -2019;
Δ = b2-4ac
Δ = 12-4·2019·(-2019)
Δ = 16305445
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{-(1)-\sqrt{16305445}}{2*2019}=\frac{-1-\sqrt{16305445}}{4038} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{16305445}}{2*2019}=\frac{-1+\sqrt{16305445}}{4038} $
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