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x^2+2x-1482=0
a = 1; b = 2; c = -1482;
Δ = b2-4ac
Δ = 22-4·1·(-1482)
Δ = 5932
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5932}=\sqrt{4*1483}=\sqrt{4}*\sqrt{1483}=2\sqrt{1483}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1483}}{2*1}=\frac{-2-2\sqrt{1483}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1483}}{2*1}=\frac{-2+2\sqrt{1483}}{2} $
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