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16t^2+8t-80=0
a = 16; b = 8; c = -80;
Δ = b2-4ac
Δ = 82-4·16·(-80)
Δ = 5184
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{5184}=72$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-72}{2*16}=\frac{-80}{32} =-2+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+72}{2*16}=\frac{64}{32} =2 $
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