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x^2+120x-1440=0
a = 1; b = 120; c = -1440;
Δ = b2-4ac
Δ = 1202-4·1·(-1440)
Δ = 20160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20160}=\sqrt{576*35}=\sqrt{576}*\sqrt{35}=24\sqrt{35}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-24\sqrt{35}}{2*1}=\frac{-120-24\sqrt{35}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+24\sqrt{35}}{2*1}=\frac{-120+24\sqrt{35}}{2} $
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