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x^2+4x=120

We move all terms to the left:

x^2+4x-(120)=0

a = 1; b = 4; c = -120;

Δ = b^{2}-4ac

Δ = 4^{2}-4·1·(-120)

Δ = 496

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{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{31}}{2*1}=\frac{-4-4\sqrt{31}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{31}}{2*1}=\frac{-4+4\sqrt{31}}{2} $

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