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x^2+2x-2808=0
a = 1; b = 2; c = -2808;
Δ = b2-4ac
Δ = 22-4·1·(-2808)
Δ = 11236
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{11236}=106$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-106}{2*1}=\frac{-108}{2} =-54 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+106}{2*1}=\frac{104}{2} =52 $
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