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5=0.8t^2
We move all terms to the left:
5-(0.8t^2)=0
We get rid of parentheses
-0.8t^2+5=0
a = -0.8; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-0.8)·5
Δ = 16
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{16}=4$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*-0.8}=\frac{-4}{-1.6} =2+0.8/1.6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-0.8}=\frac{4}{-1.6} =-2+0.8/1.6 $
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