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16t^2-120t+216=0
a = 16; b = -120; c = +216;
Δ = b2-4ac
Δ = -1202-4·16·216
Δ = 576
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-24}{2*16}=\frac{96}{32} =3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+24}{2*16}=\frac{144}{32} =4+1/2 $
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