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9t^2-15t=0
a = 9; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·9·0
Δ = 225
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{225}=15$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*9}=\frac{0}{18} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*9}=\frac{30}{18} =1+2/3 $
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