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120x^2-482x+480=0
a = 120; b = -482; c = +480;
Δ = b2-4ac
Δ = -4822-4·120·480
Δ = 1924
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1924}=\sqrt{4*481}=\sqrt{4}*\sqrt{481}=2\sqrt{481}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-482)-2\sqrt{481}}{2*120}=\frac{482-2\sqrt{481}}{240} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-482)+2\sqrt{481}}{2*120}=\frac{482+2\sqrt{481}}{240} $
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