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x^2+13x-484=0
a = 1; b = 13; c = -484;
Δ = b2-4ac
Δ = 132-4·1·(-484)
Δ = 2105
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{2105}}{2*1}=\frac{-13-\sqrt{2105}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{2105}}{2*1}=\frac{-13+\sqrt{2105}}{2} $
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