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16t^2+96t+80=0
a = 16; b = 96; c = +80;
Δ = b2-4ac
Δ = 962-4·16·80
Δ = 4096
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{4096}=64$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-64}{2*16}=\frac{-160}{32} =-5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+64}{2*16}=\frac{-32}{32} =-1 $
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