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9t^2+17t-240=0
a = 9; b = 17; c = -240;
Δ = b2-4ac
Δ = 172-4·9·(-240)
Δ = 8929
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{-(17)-\sqrt{8929}}{2*9}=\frac{-17-\sqrt{8929}}{18} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{8929}}{2*9}=\frac{-17+\sqrt{8929}}{18} $
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