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3=1+9t-5t^2
We move all terms to the left:
3-(1+9t-5t^2)=0
We get rid of parentheses
5t^2-9t-1+3=0
We add all the numbers together, and all the variables
5t^2-9t+2=0
a = 5; b = -9; c = +2;
Δ = b2-4ac
Δ = -92-4·5·2
Δ = 41
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{41}}{2*5}=\frac{9-\sqrt{41}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{41}}{2*5}=\frac{9+\sqrt{41}}{10} $
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