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5t^2+9t-11=0

a = 5; b = 9; c = -11;

Δ = b^{2}-4ac

Δ = 9^{2}-4·5·(-11)

Δ = 301

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{-(9)-\sqrt{301}}{2*5}=\frac{-9-\sqrt{301}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{301}}{2*5}=\frac{-9+\sqrt{301}}{10} $

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