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=12+120S-16S^2
We move all terms to the left:
-(12+120S-16S^2)=0
We get rid of parentheses
16S^2-120S-12=0
a = 16; b = -120; c = -12;
Δ = b2-4ac
Δ = -1202-4·16·(-12)
Δ = 15168
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{15168}=\sqrt{64*237}=\sqrt{64}*\sqrt{237}=8\sqrt{237}$$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-8\sqrt{237}}{2*16}=\frac{120-8\sqrt{237}}{32} $$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+8\sqrt{237}}{2*16}=\frac{120+8\sqrt{237}}{32} $
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