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16t^2-120t+180=0
a = 16; b = -120; c = +180;
Δ = b2-4ac
Δ = -1202-4·16·180
Δ = 2880
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-24\sqrt{5}}{2*16}=\frac{120-24\sqrt{5}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+24\sqrt{5}}{2*16}=\frac{120+24\sqrt{5}}{32} $
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