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x^2+2x-5120=0
a = 1; b = 2; c = -5120;
Δ = b2-4ac
Δ = 22-4·1·(-5120)
Δ = 20484
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20484}=\sqrt{36*569}=\sqrt{36}*\sqrt{569}=6\sqrt{569}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{569}}{2*1}=\frac{-2-6\sqrt{569}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{569}}{2*1}=\frac{-2+6\sqrt{569}}{2} $
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