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40=30t-5t^2
We move all terms to the left:
40-(30t-5t^2)=0
We get rid of parentheses
5t^2-30t+40=0
a = 5; b = -30; c = +40;
Δ = b2-4ac
Δ = -302-4·5·40
Δ = 100
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{100}=10$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-10}{2*5}=\frac{20}{10} =2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+10}{2*5}=\frac{40}{10} =4 $
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