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5t^2+39t-1=0
a = 5; b = 39; c = -1;
Δ = b2-4ac
Δ = 392-4·5·(-1)
Δ = 1541
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{-(39)-\sqrt{1541}}{2*5}=\frac{-39-\sqrt{1541}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{1541}}{2*5}=\frac{-39+\sqrt{1541}}{10} $
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