x^2+4x=120

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Solution for x^2+4x=120 equation:



x^2+4x=120
We move all terms to the left:
x^2+4x-(120)=0
a = 1; b = 4; c = -120;
Δ = b2-4ac
Δ = 42-4·1·(-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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