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5t^2=-9t+18
We move all terms to the left:
5t^2-(-9t+18)=0
We get rid of parentheses
5t^2+9t-18=0
a = 5; b = 9; c = -18;
Δ = b2-4ac
Δ = 92-4·5·(-18)
Δ = 441
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{441}=21$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-21}{2*5}=\frac{-30}{10} =-3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+21}{2*5}=\frac{12}{10} =1+1/5 $
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