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X^2+4X=360
We move all terms to the left:
X^2+4X-(360)=0
a = 1; b = 4; c = -360;
Δ = b2-4ac
Δ = 42-4·1·(-360)
Δ = 1456
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1456}=\sqrt{16*91}=\sqrt{16}*\sqrt{91}=4\sqrt{91}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{91}}{2*1}=\frac{-4-4\sqrt{91}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{91}}{2*1}=\frac{-4+4\sqrt{91}}{2} $
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