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15=-5t^2
We move all terms to the left:
15-(-5t^2)=0
We get rid of parentheses
5t^2+15=0
a = 5; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·5·15
Δ = -300
Delta is less than zero, so there is no solution for the equation
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