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9+3t-2t^2=0
a = -2; b = 3; c = +9;
Δ = b2-4ac
Δ = 32-4·(-2)·9
Δ = 81
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{81}=9$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*-2}=\frac{-12}{-4} =+3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*-2}=\frac{6}{-4} =-1+1/2 $
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