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80-5t^2=0
a = -5; b = 0; c = +80;
Δ = b2-4ac
Δ = 02-4·(-5)·80
Δ = 1600
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{1600}=40$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*-5}=\frac{-40}{-10} =+4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*-5}=\frac{40}{-10} =-4 $
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