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2x^2+x-2499=0
a = 2; b = 1; c = -2499;
Δ = b2-4ac
Δ = 12-4·2·(-2499)
Δ = 19993
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{19993}}{2*2}=\frac{-1-\sqrt{19993}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{19993}}{2*2}=\frac{-1+\sqrt{19993}}{4} $
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