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2t^2-16t-40=0
a = 2; b = -16; c = -40;
Δ = b2-4ac
Δ = -162-4·2·(-40)
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-24}{2*2}=\frac{-8}{4} =-2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+24}{2*2}=\frac{40}{4} =10 $
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