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5t^2+33t-80=0
a = 5; b = 33; c = -80;
Δ = b2-4ac
Δ = 332-4·5·(-80)
Δ = 2689
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{-(33)-\sqrt{2689}}{2*5}=\frac{-33-\sqrt{2689}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+\sqrt{2689}}{2*5}=\frac{-33+\sqrt{2689}}{10} $
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