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x^2+4x-484=0
a = 1; b = 4; c = -484;
Δ = b2-4ac
Δ = 42-4·1·(-484)
Δ = 1952
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1952}=\sqrt{16*122}=\sqrt{16}*\sqrt{122}=4\sqrt{122}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{122}}{2*1}=\frac{-4-4\sqrt{122}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{122}}{2*1}=\frac{-4+4\sqrt{122}}{2} $
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